Question

Решить) 1)38sinx=cosx 2)sinx=cosx-1

Answer 1

$38\sin x=\cos x\,|:\cos x \\ 38tg x=1 \\ tg x= \frac{1}{38} \\ x=arctg \frac{1}{38}+ \pi n, n \in Z$

$\sin x=\cos x-1 \\ \cos x-\sin x=1 \\ \sqrt{2} \sin(x- \frac{ \pi }{4} )=1 \\ \sin(x- \frac{ \pi }{4} )= \frac{1}{ \sqrt{2} } \\ x-\frac{ \pi }{4}=(-1)^k\cdot \frac{ \pi }{4}+ \pi k, k \in Z \\ x=(-1)^k\cdot \frac{ \pi }{4}+\frac{ \pi }{4}+ \pi k, k \in Z$