Question

[Thùy Oanh Nguyễn]

Giải pt sau:

A. (Sinx+1)(sinx-√2)=0

B.2sinxcosx=1

C. 4sinxcosxcos2x+1=0

D. Sin⁴x-cos⁴x=0

E. (Sinx+1)(2cos2x-√2)

F. Sin2x=cos⁴x/2-sin⁴x/2

Answer 1

[Nguyễn Việt Lâm]

d.

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

\(\Leftrightarrow sin^2x-cos^2x=0\)

\(\Leftrightarrow-cos2x=0\)

\(\Leftrightarrow2x=\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

e. Đề thiếu

f.

\(\Leftrightarrow sin2x=\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)\)

\(\Leftrightarrow sin2x=cos^2\frac{x}{2}-sin^2\frac{x}{2}\)

\(\Leftrightarrow sin2x=cosx\)

\(\Leftrightarrow sin2x=sin\left(\frac{\pi}{2}-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

a.

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\sqrt{2}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)

b.

\(\Leftrightarrow sin2x=1\)

\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)

c.

\(\Leftrightarrow2sin2x.cos2x=-1\)

\(\Leftrightarrow sin4x=-1\)

\(\Leftrightarrow4x=-\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)