Question

[Nguyễn Kiều Anh]

Giải các phương trình sau:

a, sinx+cosx+1+sin2x+cos2x=0

b, sinx(1+cos2x)+sin2x=1+cos2x

c, \(\frac{1}{sinx}+\frac{1}{sin\left(x-\frac{3\pi}{2}\right)}=4sin\left(\frac{7\pi}{4}-x\right)\)

d, sin⁴x+cos⁴x=\(\frac{7}{8}cot\left(x+\frac{\pi}{3}\right)cot\left(\frac{\pi}{6}-x\right)\)

Answer 1

[Nguyễn Kiều Anh]

@Nguyễn Việt Lâm giúp em với ạ

Answer 2

[Nguyễn Việt Lâm]

a.

\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)

\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)