Question

Решите уравнение: 1) $sin^{2}$x+4sin x+3=0
2) $cos^{2}$x-2cos x-3=0

Answer 1

$1) sin ^{2} x+4sinx+3=0\\sinx=t\\-1 \leq t \leq 1 \\ t ^{2} +4t+3=0\\D=16-12=4 \\ \sqrt{D} =2\\ t_{1} = \frac{-4+2}{2} =-1 \\ t _{2} = \frac{-4-2}{2} =-3\\sinx=-1\\x=- \frac{ \pi }{2} +2 \pi n$
n ∈ Z 

$2) cos ^{2} x-2cosx-3=0\\cosx=t \\ -1 \leq t \leq 1\\t ^{2} -2t-3=0\\D=4+12=16\\ \sqrt{D} =4\\ t_{1} = \frac{2+4}{2} =3\\t _{2} = \frac{2-4}{2} =-1\\cosx=-1\\x= \pi +2 \pi k$

k ∈ Z