Question

Giải pt

a, \(sin\left(3x+\frac{\pi}{3}\right)+sin\left(\frac{4\pi}{5}-3x\right)=\sqrt{3}\)

b, \(2tanx.cosx+1=2cosx+tanx\)

c, \(tanx+tan2x=tan3x\)

d, cos2x + sin2x = \(\frac{\sqrt{6}}{2}\)

e, \(2tan^2x+3=\frac{3}{cosx}\)

f. \(sin^24x+sin^23x=sin^22x+sin^2x\)

thanks youuuuu mấy bài này khó quá mình suy nghĩ mãi hong ra cảm ơn trước nhaa

Answer 1

Answer 2

a.

\(\Leftrightarrow2sin\frac{17\pi}{30}cos\left(3x-\frac{7\pi}{30}\right)=\sqrt{3}\)

\(\Leftrightarrow cos\left(3x-\frac{7\pi}{30}\right)=\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}\)

Đặt \(\frac{\sqrt{3}}{2sin\left(\frac{17\pi}{30}\right)}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow cos\left(3x-\frac{7\pi}{30}\right)=cosa\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{7\pi}{30}=a+k2\pi\\3x-\frac{7\pi}{30}=-a+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{90}+\frac{a}{3}+\frac{k2\pi}{3}\\x=\frac{7\pi}{30}-\frac{a}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)

Chắc bạn ghi sai đề, con số \(\frac{4\pi}{3}\) sẽ hợp lý hơn con số \(\frac{4\pi}{5}\) rất nhiều

Answer 3

b.

ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow2tanx.cosx-tanx+1-2cosx=0\)

\(\Leftrightarrow tanx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left(2cosx-1\right)\left(tanx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\tanx=1\end{matrix}\right.\)

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

Answer 4

c.

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne0\\cos3x\ne0\end{matrix}\right.\)

\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}=\frac{sin3x}{cos3x}\)

\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}=\frac{sin3x}{cos3x}\)

\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}=\frac{sin3x}{cos3x}\)

\(\Leftrightarrow sin3x\left(\frac{cos3x-cosx.cos2x}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow\frac{1}{2}sin3x\left(\frac{2cos3x-cos3x-cosx}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{cos3x-cosx}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow-2sin3x\left(\frac{sin2x.sinx}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{sin^2x}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x=0\Leftrightarrow x=\frac{k\pi}{3}\) (sin3x đã bao hàm luôn sinx nên ko cần xét \(sin^2x=0\))

Answer 5

f.

\(\frac{1}{2}-\frac{1}{2}cos8x+\frac{1}{2}-\frac{1}{2}cos6x=\frac{1}{2}-\frac{1}{2}cos4x+\frac{1}{2}-\frac{1}{2}cos2x\)

\(\Leftrightarrow cos8x+cos6x=cos4x+cos2x\)

\(\Leftrightarrow2cos7x.cosx=2cos3x.cosx\)

\(\Leftrightarrow cosx.\left(cos7x-cos3x\right)=0\)

\(\Leftrightarrow-2cosx.sin5x.sin2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin5x=0\\sin2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{5}\\x=\frac{k\pi}{2}\end{matrix}\right.\)