Pregunta 1. Encuentra la solución general de las siguientes ecuaciones:
(i) sen θ = 1/2
Solución:
Se nos da,
=> sen θ = 1/2
=> sen θ = sen π/6
Sabemos que si sen θ = sen a, la solución general está dada por, θ = nπ + (−1) n a; n ∈ Z.
=>θ = nπ + (−1) n (π/6) ; norte ∈ Z
(ii) cos θ = −√3/2
Solución:
Se nos da,
=> cos θ = −√3/2
=> cos θ = cos (π + π/6)
=> cos θ = cos (7π/6)
Sabemos que si cos θ = cos a, la solución general está dada por, θ = 2nπ ± a ; n ∈ Z.
=> θ = 2nπ ± (7π/6) ; norte ∈ Z
(iii) cosec θ = −√2
Solución:
Se nos da,
=> cosec θ = −√2
=> 1/sen θ = −√2
=> sen θ = −1/√2
=> sen θ = −sen (π/4)
Como sen (−θ) = − sen θ, tenemos,
=> sen θ = sen (−π/4)
Sabemos que si sen θ = sen a, la solución general está dada por, θ = nπ + (−1) n a ; n ∈ Z.
=> θ = nπ + (−1) n+1 (π/4) ; norte ∈ Z
(iv) segundo θ = √2
Solución:
Se nos da,
=> segundo θ = √2
=> 1/seg θ = √2
=> cos θ = 1/√2
=> cos θ = cos π/4
Sabemos que si cos θ = cos a, la solución general está dada por, θ = 2nπ ± a ; n ∈ Z.
=> θ = 2nπ ± (π/4) ; norte ∈ Z
(v) tan θ = −1/√3
Solución:
Se nos da,
=> bronceado θ = −1/√3
=> tan θ = −tan (π/6)
Como tan (−θ) = − tan θ, tenemos,
=> tan θ = tan (−π/6)
Sabemos que si tan θ = tan a, la solución general está dada por, θ = nπ + a ; n ∈ Z.
=> θ = nπ − (π/6) ; norte ∈ Z
(vi) √3 seg θ = 2
Solución:
Se nos da,
=> √3 seg θ = 2
=> √3 (1/cos θ) = 2
=> cos θ = √3/2
=> cos θ = cos (π/6)
Sabemos que si cos θ = cos a, la solución general está dada por, θ = 2nπ ± a; n ∈ Z.
=> θ = 2nπ ± (π/6) ; norte ∈ Z
Pregunta 2. Encuentra la solución general de las siguientes ecuaciones:
(i) sen 2 θ = √3/2
Solución:
Se nos da,
=> sen 2θ = √3/2
=> sen 2θ = sen (π/3)
Sabemos que si sen θ = sen a, la solución general está dada por, θ = nπ + (−1) n a; n ∈ Z.
=> 2θ = nπ + (−1) n (π/3)
=>θ = nπ/2 + (−1) n (π/6) ; norte ∈ Z
(ii) cos 3 θ = 1/2
Solución:
Se nos da,
=> cos 3θ = 1/2
=> cos 3θ = cos (π/3)
Sabemos que si cos θ = cos a, la solución general está dada por, θ = 2nπ ± a; n ∈ Z.
=> 3θ = 2nπ ± (π/3)
=>θ = 2nπ/3 ± (π/9) ; norte ∈ Z
(iii) sen 9 θ = sen θ
Solución:
Se nos da,
=> sen 9θ = sen θ
=> sen 9θ – sen θ = 0
=> 2 cos 5θ sen 4θ = 0
=> cos 5θ = 0 o sen 4θ = 0
Sabemos, si cos θ = 0, entonces θ = (2n+1)π/2 y si sen θ = 0, entonces θ = nπ donde n ∈ Z.
=> 5θ = (2n+1)π/2 o 4θ = nπ
=> θ = (2n+1)π/10 o θ = nπ/4, n ∈ Z
(iv) sen 2θ = cos 3θ
Solución:
Se nos da,
=> sen 2θ = cos 3θ
=> cos 3θ = sen 2θ
=> cos 3θ = cos (π/2 − 2θ)
Sabemos que si cos θ = cos a, la solución general está dada por, θ = 2nπ ± a ; n ∈ Z.
=> 3θ = 2nπ ± (π/2 − 2θ)
=> 3θ = 2nπ + π/2 − 2θ o 3θ = 2nπ − π/2 + 2θ
=> 5θ = 2nπ + π/2 o θ = 2nπ − π/2
=> 5θ = (4n+1) (π/2) o θ = (4n−1) (π/2)
=> θ = (4n+1)π/10 o θ = (4n−1)π/2, n ∈ Z
(v) tan θ + cot 2θ = 0
Solución:
Se nos da,
=> tan θ + cot 2θ = 0
=> cuna 2θ = − tan θ
=> bronceado 2θ = − cuna θ
=> tan 2θ = − tan (π/2 − θ)
Como tan (−θ) = − tan θ, tenemos,
=> tan 2θ = tan (θ − π/2)
Sabemos que si tan θ = tan a, la solución general está dada por, θ = nπ + a ; n ∈ Z.
=> 2θ = nπ + θ − π/2
=>θ = nπ − π/2 ; norte ∈ Z
(vi) tan 3θ = cot θ
Solución:
Se nos da,
=> tan 3θ = cuna θ
=> tan 3θ = tan (π/2 − θ)
Sabemos que si tan θ = tan a, la solución general está dada por, θ = nπ + a ; n ∈ Z.
=> 3θ = nπ + π/2 − θ
=> 4θ = nπ + π/2
=> θ = nπ/4 + π/8 ; norte ∈ Z
(vii) tan 2θ tan θ = 1
Solución:
Se nos da,
=> tan 2θ tan θ = 1
=> tan 2θ = cuna θ
=> tan 2θ = tan (π/2 − θ)
Sabemos que si tan θ = tan a, la solución general está dada por, θ = nπ + a ; n ∈ Z.
=> 2θ = nπ + π/2 − θ
=> 3θ = nπ + π/2
=> θ = nπ/3 + π/6 ; norte ∈ Z
(viii) tan mθ + cot nθ = 0
Solución:
Se nos da,
=> tan mθ + cot nθ = 0
=> sen mθ/cos mθ + cos nθ/sen nθ = 0
=> sen mθ sen nθ + cos nθ cos mθ = 0
=> cos (m−n)θ = 0
Sabemos que si cos θ = 0, entonces θ = (2k+1)π/2 donde k ∈ Z.
=> (m−n)θ = (2k+1)π/2
=> θ = (2k+1)π/2(m−n) ; norte ∈ Z
(ix) tan pθ = cot qθ
Solución:
Se nos da,
=> tan pθ = cot qθ
=> tan pθ = tan (π/2 − qθ)
Sabemos que si tan θ = tan a, la solución general está dada por, θ = nπ + a ; n ∈ Z.
=> pθ = nπ + π/2 − qθ
=> (p + q)θ = (2n+1)π/2
=> θ = (2n+1)π/2(p + q) ; norte ∈ Z
(x) sen 2x + cos x = 0
Solución:
Se nos da,
=> sen 2x + cos x = 0
=> cosx = − sen 2x
=> cos x = cos (π/2 + 2x)
Sabemos que si cos θ = cos a, la solución general está dada por, θ = 2nπ ± a ; n ∈ Z.
=> x = 2nπ ± (π/2 + 2x)
=> x = 2nπ + π/2 + 2x o x = 2nπ − π/2 − 2x
=> x = −(4n+1)π/2 o 3x = (4n−1)π/2
=> x = −(4n+1)π/2 o 3x = (4n−1)π/2
=> x = (4m−1)π/2, donde m = −n o x = (4n−1)π/6, n ∈ Z
(xi) sen θ = tan θ
Solución:
Se nos da,
=> sen θ = tan θ
=> sen θ = sen θ/cos θ
=> sen θ (cos θ − 1) = 0
=> sen θ = 0 o cos θ = 1
=> sen θ = 0 o cos θ = cos 0
Sabemos, si cos θ = 0, entonces θ = (2n+1)π/2 y si sen θ = 0, entonces θ = nπ donde n ∈ Z.
=> θ = nπ o θ = 2nπ, n ∈ Z
(xii) sen 3x + cos 2x = 0
Solución:
Se nos da,
=> sen 3x + cos 2x = 0
=> cos 2x = −sen 3x
=>cos 2x = cos (π/2 + 3x)
Sabemos que si cos θ = cos a, la solución general está dada por, θ = 2nπ ± a ; n ∈ Z.
=> 2x = 2nπ ± (π/2 + 3x)
=> 2x = 2nπ + π/2 + 3x o 2x = 2nπ − π/2 − 3x
=> −x = 2nπ + π/2 o 5x = 2nπ − π/2
=> x = 2mπ − π/2, donde m = −n o x = (2nπ − π/2)/5
=> x = (4m−1)π/2, donde m = −n o x = (4n−1)π/10, n ∈ Z
Pregunta 3. Resuelve las siguientes ecuaciones:
(i) sen 2 θ − cos θ = 1/4
Solución:
Tenemos,
=> sen 2 θ − cos θ = 1/4
=> 1 − cos 2 θ − cos θ = 1/4
=> cos 2 θ + cos θ − 3/4 = 0
=> 4 cos 2 θ + 4 cos θ − 3 = 0
=> 4 cos 2 θ + 6 cos θ − 2 cos θ − 3 = 0
=> 2 cos θ (2 cos + 3) − (2 cos θ + 3) = 0
=> (2 cos θ − 1) (2 cos θ + 3) = 0
=> cos θ = 1/2 o cos θ = −3/2
Ignorando cos θ = −3/2 como −1 ≤ cosθ ≤ 1. Entonces, tenemos, cos θ = 1/2.
=> cos θ = cos π/3
=>θ = 2nπ ± π/3 ; norte ∈ Z
(ii) 2cos 2 θ − 5cos θ + 2 = 0
Solución:
Se nos da,
=> 2cos 2 θ − 5cos θ + 2 = 0
=> 2cos 2 θ − 4cos θ − cos θ + 2 = 0
=> 2 cos θ (cos θ − 2) − (cos θ − 2) = 0
=> (2cos θ − 1) (cos θ − 2) = 0
=> cos θ = 1/2 o cos θ = 2
Ignorando cos θ = 2 como −1 ≤ cosθ ≤ 1. Entonces, tenemos, cos θ = 1/2.
=> cos θ = cos π/3
=> θ = 2nπ ± π/3 ; norte ∈ Z
(iii) 2sen 2x + √3cos x + 1 = 0
Solución:
Se nos da,
=> 2sen 2x + √3cos x + 1 = 0
=> 2 (1 − cos 2 x) + √3 cos x + 1 = 0
=> 2 cos 2 x − √3 cos x − 3 = 0
=> 2cos 2 x − 2√3cos x + √3cos x − 3 = 0
=> 2cosx (cos x − √3) + √3(cos x − √3) = 0
=> (2cosx + √3) (cos x − √3) = 0
=> x = −√3/2 o x = √3
Ignorando cos x = √3 como −1 ≤ cosθ ≤ 1. Entonces, tenemos, cos x = −√3/2.
=> cos x = − cos π/6
=> cos x = cos (π − π/6)
=> cos x = cos (5π/6)
=> x = 2nπ ± 5π/6 ; norte ∈ Z
(iv) 4sen 2 θ − 8cos θ + 1 = 0
Solución:
Se nos da,
=> 4sen 2 θ − 8cos θ + 1 = 0
=> 4 (1 − cos 2 θ) − 8 cos θ + 1 = 0
=> 4 porque 2 θ + 8 porque θ − 5 = 0
=> 4 porque 2 θ + 10 porque θ − 2 porque θ − 5 = 0
=> 2cos θ (2cos θ +5) − (2cos θ +5) = 0
=> (2cos θ − 1) (2cos θ +5) = 0
=> cos θ = 1/2 o cos θ = −5/2
Ignorando cos x = −5/2 como −1 ≤ cosθ ≤ 1. Entonces, tenemos, cos x = 1/2.
=> cos x = cos π/3
=> x = 2nπ ± π/3 ; norte ∈ Z
(v) bronceado 2 x + (1 − √3) bronceado x − √3 = 0
Solución:
Se nos da,
=> bronceado 2 x + (1 − √3) bronceado x − √3 = 0
=> tan 2 x + tan x − √3 tan x − √3 = 0
=> tan x (tan x + 1) − √3 (tan x + 1) = 0
=> (bronceado x − √3) (bronceado x + 1) = 0
=> tan x = √3 o tan x = −1
=> tan x = tan π/3 o tan x = −tan π/4
=> tan x = tan π/3 o tan x = tan (−π/4)
=> x = nπ + π/3 o x = nπ − π/4, n ∈ Z
(vi) 3 cos 2 θ − 2√3 sen θ cos θ − 3 sen 2 θ = 0
Solución:
Tenemos,
=> 3 cos 2 θ − 2√3 sen θ cos θ − 3 sen 2 θ = 0
=> √3 cos 2 θ − 2 sen θ cos θ − √3 sen 2 θ = 0
=> √3 cos 2 θ + sen θ cos θ − 3 sen θ cos θ − √3 sen 2 θ = 0
=> cos θ (√3 cos θ + sen θ) − √3 sen θ (√3 cos θ + sen θ) = 0
=> (√3cos θ + sen θ) (cos θ − √3sen θ) = 0
=> tan θ = −√3 o tan θ = 1/√3
=> tan θ = − tan π/3 o tan θ = tan π/6
=> tan θ = tan (−π/3) o tan θ = tan π/6
=> θ = nπ − π/3 o θ = nπ + π/6, n ∈ Z
(vii) cos 4θ = cos 2θ
Solución:
Tenemos,
=> cos 4θ = cos 2θ
=> cos 4θ − cos 2θ = 0
=> 2 sen 3θ sen θ = 0
=> sen 3θ = 0 o sen θ = 0
=> 3θ = nπ o θ = nπ
=> θ = nπ/3 o θ = nπ, n ∈ Z
Pregunta 4. Resuelve las siguientes ecuaciones:
(i) cos θ + cos 2θ + cos 3θ = 0
Solución:
Se nos da,
=> cos 2θ + (cos θ + cos 3θ) = 0
=> cos 2θ + 2 cos 2θ cos θ = 0
=> cos 2θ (1 + 2 cos θ) = 0
=> cos 2θ = 0 o cos θ = −1/2
=> 2θ = (2n+1)π/2 o cos θ = cos (π − π/3)
=> θ = (2n+1)π/4 o θ = 2nπ ± (2π/3), n ∈ Z
(ii) cos θ + cos 3θ − cos 2θ = 0
Solución:
Se nos da,
=> cos θ + cos 3θ − cos 2θ = 0
=> 2cos 2θ cos θ − cos 2θ = 0
=> cos 2θ (2 cos θ − 1) = 0
=> cos 2θ = 0 o cos θ = 1/2
=> 2θ = (2n+1)π/2 o cos θ = cos π/3
=> θ = (2n+1)π/4 o θ = 2nπ ± (π/3), n ∈ Z
(iii) sen θ + sen 5θ = sen 3θ
Solución:
Se nos da,
=> sen θ + sen 5θ = sen 3θ
=> 2sen 3θ cos 2θ − sen 3θ = 0
=> sen 3θ (2cos 2θ − 1) = 0
=> sen 3θ = 0 o cos 2θ = 1/2
=> sen 3θ = 0 o cos 2θ = cos π/3
=> 3θ = nπ o 2θ = 2nπ ± (π/3)
=> θ = nπ/3 o θ = nπ ± (π/6), n ∈ Z
(iv) cos θ cos 2θ cos 3θ = 1/4
Solución:
Se nos da,
=> cos θ cos 2θ cos 3θ = 1/4
=> 2cos θ cos 3θ cos 2θ = 1/2
=> (cos 4θ + cos 2θ) cos 2θ = 1/2
=> (2cos 2 2θ − 1 + cos 2θ) cos 2θ = 1/2
=> 2cos 3 2θ − cos 2θ + cos 2 2θ = 1/2
=> 4cos 3 2θ − 2cos 2θ + 2cos 2 2θ − 1 = 0
=> 2cos 2 2θ (2cos 2θ + 1) − (2cos 2θ + 1) = 0
=> (2cos 2 2θ − 1) (2cos 2θ + 1) = 0
=> 2cos 2 2θ − 1 = 0 o 2cos 2θ + 1 = 0
=> cos 4θ = 0 o cos 2θ = −1/2
=> 4θ = (2n+1)π/2 o cos 2θ = cos 2π/3
=> θ = (2n+1)π/8 o 2θ = 2nπ ± (2π/3)
=> θ = (2n+1)π/8 o 2θ = nπ ± (π/3), n ∈ Z
(v) cos θ + sen θ = cos 2θ + sen 2θ
Solución:
Se nos da,
=> cos θ + sen θ = cos 2θ + sen 2θ
=> cos θ − cos 2θ = sen 2θ − sen θ
=> 2 sen 3θ/2 sen θ/2 = 2 cos 3θ/2 sen θ/2
=> 2 sen θ/2 (sen 3θ/2 − cos 3θ/2) = 0
=> sen θ/2 = 0 o (sen 3θ/2 − cos 3θ/2) = 0
=> θ/2 = nπ o tan 3θ/2 = 1
=> θ = 2nπ o tan 3θ/2 = tan π/4
=> θ = 2nπ o 3θ/2 = nπ ± π/4
=> θ = 2nπ o θ = 2nπ/3 ± π/6, n ∈ Z
(vi) sen θ + sen 2θ + sen 3θ = 0
Solución:
Se nos da,
=> (sen θ + sen 3θ) + sen 2θ = 0
=> 2sen 2θ cos θ + sen 2θ = 0
=> sen 2θ (2 cos θ + 1) = 0
=> sen 2θ = 0 o 2cos θ + 1 = 0
=> 2θ = nπ o cos θ = −1/2
=> θ = nπ/2 o cos θ = cos 2π/3
=> θ = nπ/2 o θ = 2nπ ± (2π/3), n ∈ Z
(vii) sen x + sen 2x + sen 3x + sen 4x = 0
Solución:
Se nos da,
=> sen x + sen 2x + sen 3x + sen 4x = 0
=> (sen x + sen 3x) + (sen 2x + sen 4x) = 0
=> 2 sen 2x cos x + 2 sen 3x cos x = 0
=> 2cos x (sen 2x + sen 3x) = 0
=> (2cos x) (2sen 5x/2) (cos x/2) = 0
=> cos x = 0 o sen 5x/2 = 0 o cos x/2 = 0
=> x = (2n+1)π/2 o 5x/2 = nπ o cos x/2 = 0
=> x = (2n+1)π/2 o x = 2nπ/5 o x/2 = (2n+1)π/2
=> x = (2n+1)π/2 o x = 2nπ/5 o x = (2n+1)π, n ∈ Z
(viii) sen 3θ − sen θ = 4 cos 2 θ − 2
Solución:
Se nos da,
=> sen 3θ − sen θ = 4 cos 2 θ − 2
=> 2 cos 2θ sen θ = 2 (2 cos 2 θ − 1)
=> 2 cos 2θ sen θ = 2 cos 2θ
=> 2 cos 2θ (sen θ − 1) = 0
=> cos 2θ = 0 o (sen θ − 1) = 0
=> 2θ = (2n+1)π/2 o sen θ = 1
=> θ = (2n+1)π/4 o sen θ = sen π/2
=> θ = (2n+1)π/4 o θ = nπ + (−1) n (π/2), n ∈ Z
(ix) sen 2x− sen 4x + sen 6x = 0
Solución:
Se nos da,
=> sen 2x− sen 4x + sen 6x = 0
=> 2 sen 4x cos 2x − sen 4x = 0
=> 2 sen 4x (cos 2x − 1) = 0
=> sen 4x = 0 o cos 2x = 1/2
=> 4x = nπ o cos 2x = cos π/3
=> x = nπ/4 o 2x = 2nπ ± (π/3)
=> x = nπ/4 o x = nπ ± (π/6), n ∈ Z
Pregunta 5. Resuelve las siguientes ecuaciones:
(i) tan x + tan 2x + tan 3x = 0
Solución:
Se nos da,
=> tan x + tan 2x + tan 3x = 0
=> (tan x + tan 2x) + tan (2x+x) = 0
=> (tan x + tan 2x) + (tan x + tan 2x)/(1 − tan x tan 2x) = 0
=> (tan x + tan 2x) [1 + 1/(1 − tan x tan 2x)] = 0
=> (tan x + tan 2x) (2 − tan x tan 2x) = 0
=> (tan x + tan 2x) = 0 o (2 − tan x tan 2x) = 0
=> tan x = −tan 2x o tan x tan 2x = 2
=> tan x = tan (−2x) o tan x [2tan x/(1−tan 2 x)] = 2
=> x = nπ − 2x o 2tan 2 x = 2 − tan 2 x
=> 3x = nπ o 4tan 2 x = 2
=> x = nπ/3 o tan 2 x = 1/2
=> x = nπ/3 o tan x = 1/√2 o tan x = −1/√2
=> x = nπ/3 o tan x = tan π/4 o tan x = tan (−π/4)
=> x = nπ/3 o x = nπ + π/4 o x = nπ − π/4
=> x = nπ/3 o x = nπ ± π/4, n ∈ Z
(ii) tan θ + tan 2θ = tan 3θ
Solución:
Se nos da,
=> bronceado θ + bronceado 2θ = bronceado 3θ
=> (tan θ + tan 2θ) − tan (2θ+θ) = 0
=> (tan θ + tan 2θ) − (tan θ + tan 2θ)/(1 − tan θ tan 2θ) = 0
=> (tan θ + tan 2θ) [1 − 1/(1 − tan θ tan 2θ)] = 0
=> (tan θ + tan 2θ) (−tan θ tan 2θ) = 0
=> (tan θ + tan 2θ) = 0 o −tan θ = 0 o tan 2θ = 0
=> tan θ = tan (−2θ) o tan θ = 0 o tan 2θ = 0
=> θ = nπ − 2θ o θ = nπ o 2θ = nπ
=> θ = nπ/3 o θ = nπ o θ = nπ/2, n ∈ Z
(iii) tan 3θ + tan θ = 2tan 2θ
Solución:
Se nos da,
=> bronceado 3θ + bronceado θ = 2 bronceado 2θ
=> tan 3θ − tan 2θ = tan 2θ − tan θ
=> 2 sen 2 θ sen 2θ = 0
=> sen θ = 0 o sen 2θ = 0
=> θ = nπ o θ = nπ/2, n ∈ Z
Pregunta 6. Resuelve las siguientes ecuaciones:
(i) sen θ + cos θ = √2
Solución:
Se nos da,
=> sen θ + cos θ = √2
=> (1/√2) sen θ + (1/√2) cos θ = 1
=> sen π/4 sen θ + cos π/4 cos θ = 1
=> coseno (θ − π/4) = coseno 0
=> θ − π/4 = 2nπ
=> θ = 2nπ + π/4
=> θ = (8n+1)π/4, n ∈ Z
(ii) √3 cos θ + sen θ = 1
Solución:
Se nos da,
=> √3 cos θ + sen θ = 1
=> (√3/2) cos θ + (1/2) sen θ = 1/2
=> cos π/6 cos θ + sen π/6 sen θ = 1/2
=> coseno (θ − π/6) = coseno π/3
=> θ − π/6 = 2nπ ± π/3
=> θ = 2nπ ± π/3 + π/6
=> θ = 2nπ + π/3 − π/6 o θ = 2nπ − π/3 + π/6
=> θ = 2nπ + π/2 o θ = 2nπ − π/6
=> θ = (4n+1)π/2 o θ = (12n−1)π/6, n ∈ Z
(iii) sen θ + cos θ = 1
Solución:
Se nos da,
=> sen θ + cos θ = 1
=> (1/√2) cos θ + (1/√2) sen θ = 1/√2
=> cos π/4 cos θ + sen π/4 sen θ = 1/√2
=> coseno (θ − π/4) = coseno π/4
=> θ − π/4 = 2nπ ± π/4
=> θ = 2nπ ± π/4 + π/4
=> θ = 2nπ + π/2 o θ = 2nπ, n ∈ Z
(iv) cosec θ = 1 + cot θ
Solución:
Se nos da,
=> cosec θ = 1 + cot θ
=> 1/sen θ = 1 + cos θ/sen θ
=> sen θ + cos θ = 1
=> (1/√2) cos θ + (1/√2) sen θ = 1/√2
=> cos π/4 cos θ + sen π/4 sen θ = 1/√2
=> coseno (θ − π/4) = coseno π/4
=> θ − π/4 = 2nπ ± π/4
=> θ = 2nπ ± π/4 + π/4
=> θ = 2nπ + π/2 o θ = 2nπ, n ∈ Z
(v) (√3 − 1) cos θ + (√3 + 1) sen θ = 2
Solución:
Se nos da,
=> (√3 − 1) cos θ + (√3 + 1) sen θ = 2
=> (√3 − 1) cos θ/2√2 + (√3 + 1) sen θ/2√2 = 2
=> sen (θ + tan -1 (√3 − 1)/(√3 + 1)) = sen π/4
=> θ = 2nπ + π/3 o θ = 2nπ − π/6, n ∈ Z
Pregunta 7. Resuelve las siguientes ecuaciones:
(i) cuna x + bronceado x = 2
Solución:
Se nos da,
=> cuna x + bronceado x = 2
=> cos x/sen x + sen x/cos x = 2
=> (cos 2 x + sen 2 x)/sen x cos x = 2
=> 2 sen x cos x = 1
=> sen 2x = 1
=> sen 2x = sen π/2
=> 2x = (2n+1)π/2
=> x = (2n+1)π/4, norte ∈ Z
(ii) 2 sen 2 θ = 3 cos θ, 0 ≤ θ ≤ 2π
Solución:
Se nos da,
=> 2 sen 2 θ = 3 cos θ
=> 2 (1 − cos 2 θ) − 3 cos θ = 0
=> 2 porque 2 θ + 3 porque θ − 2 = 0
=> 2 porque 2 θ + 4 porque θ − porque θ − 2 = 0
=> 2 cos θ (cos + 2) − (cos θ + 2) = 0
=> (2 cos θ − 1) (cos θ + 2) = 0
=> cos θ = 1/2 o cos θ = −2
Ignorando cos θ =−2 como −1 ≤ cos θ ≤ 1. Entonces, tenemos, cos θ = 1/2.
=> cos θ = cos π/3
=> θ = 2nπ ± π/3, n ∈ Z
(iii) seg x cos 5x +1 = 0, 0 ≤ x ≤ π/2
Solución:
Se nos da,
=> seg x cos 5x +1 = 0
=> (cos5x + cos x)/cos x = 0
=> 2 cos 3x cos 2x = 0
=> cos 3x = 0 o cos 2x = 0
=> 3x = (2n+1)π/2 o 2x = (2n+1)π/2
=> x = (2n+1)π/6 o x = (2n+1)π/4, n ∈ Z
(iv) 5 cos 2 θ + 7 sen 2 θ − 6 = 0
Solución:
Se nos da,
=> 5 cos 2 θ + 7 sen 2 θ − 6 = 0
=> 5 (1 − sen 2 θ) + 7 sen 2 θ − 6 = 0
=> 2 sen 2 θ+ 5 − 6 = 0
=> 2 sen 2 θ = 1
=> sen θ = ±(1/√2)
=> θ = nπ ± π/4, n ∈ Z
(v) sen x − 3 sen 2x + sen 3x = cos x − 3 cos 2x + cos 3x
Solución:
Se nos da,
=> sen x − 3 sen 2x + sen 3x = cos x − 3 cos 2x + cos 3x
=> (sen x + sen 3x) − 3 sen 2x − (cos x + cos 3x) + 3 cos 2x = 0
=> 2 sen 2x cos x − 3 sen 2x − 2 cos 2x cos x + 3 cos 2x = 0
=> sen 2x (2 cos x − 3) − cos 2x (2 cos x − 3) = 0
=> (sen 2x − cos 2x) (2 cos x − 3) = 0
=> sen 2x = cos 2x o cos x = 3/2
Ignorando cos x = 3/2 como −1 ≤ cos x ≤ 1. Entonces, tenemos, sen 2x = cos 2x.
=> bronceado 2x = 1
=> tan 2x = tan π/4
=> 2x = nπ + π/4
=> x = nπ/2 + π/8, n ∈ Z
Publicación traducida automáticamente
Artículo escrito por gurjotloveparmar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA