Question

Решите уравнение cos^2x+cosx=1/2-sin^2x

Answer 1

$Cos^{2}x+Cosx=\frac{1}{2}-Sin^{2}x\\\\(Cos^{2}x+Sin^{2}x) + Cosx =\frac{1}{2}\\\\1+Cosx=\frac{1}{2}\\\\Cosx=-\frac{1}{2}\\\\x=\pm arc Cos(-\frac{1}{2})+2\pi n,n\in Z\\\\x=\pm (\pi-arc Cos\frac{1}{2})+2\pi n,n\in Z\\\\x=\pm(\pi -\frac{\pi }{3})+2\pi n,n\in Z\\\\x=\pm\frac{2\pi }{3}+2\pi n,n\in Z\\\\Otvet:\boxed{\pm\frac{2\pi }{3}+2\pi n,n\in Z}$

Answer 2

cos²x+cosx=1/2-sin²x

-1/2+cosx+cos²x+sin²x=0

1/2(-1+2cosx+2cos²x+2sin²x)=0

-1+2cosx+2cos²x+2sin²x=0

2cosx+1=0

2cosx= -1

cosx= -1/2

x= 2πn₁+2π/3,  n₁ ∈ Z