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July 2022
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2sin((3p/2)-x)cos((p/2)-x)=sqrt(2)cosx
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Verified answer
2sin(3π/2 -x)*cos(π/2 -x) =√2 *cosx ;
-2cosx*sinx =
√2 *cosx ;
2 cosx*sinx +
√2 *cosx =0 ;
2 cosx*(sinx +
√2 / 2) = 0 ;
a)
cosx = 0 ⇒x =π/2 +π*n ,
n∈Z .
b) sinx +
√2 / 2 =0 ;
sinx = -
√2 / 2 ;
x = (-1)^(n+1)* (π/4) +π*n ,
n∈Z ..
ответ : {
π/2 +π*n ; (-1)^(n+1)* (π/4) +π*n ,
n∈Z }
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Answers & Comments
Verified answer
2sin(3π/2 -x)*cos(π/2 -x) =√2 *cosx ;-2cosx*sinx = √2 *cosx ;
2 cosx*sinx + √2 *cosx =0 ;
2 cosx*(sinx + √2 / 2) = 0 ;
a) cosx = 0 ⇒x =π/2 +π*n , n∈Z .
b) sinx + √2 / 2 =0 ;
sinx = - √2 / 2 ;
x = (-1)^(n+1)* (π/4) +π*n , n∈Z ..
ответ : { π/2 +π*n ; (-1)^(n+1)* (π/4) +π*n , n∈Z }