\(sina+sinb+sinc+3=0\)

\(\Leftrightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)=0\)

Do \(\left\{{}\begin{matrix}sina\ge-1\\sinb\ge-1\\sinc\ge-1\end{matrix}\right.\) ;\(\forall a;b;c\)

\(\Rightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)\ge0\)

Dấu "=" xảy ra khi và chỉ khi \(sina=sinb=sinc=-1\)

\(\Rightarrow cosa=cosb=cosc=0\Rightarrow cosa+cosb+cosc+10=10\)

b/ \(sinx=1-sin^2x\Rightarrow sinx=cos^2x\)

\(\Rightarrow sin^2x=cos^4x\Rightarrow1-cos^2x=cos^4x\)

\(\Rightarrow cos^4x+cos^2x=1\Rightarrow\left(cos^4x+cos^2x\right)^2=1\)

\(\Rightarrow cos^8x+2cos^6x+cos^4x=1\)

Các bước biến đổi. Bạn tự tìm kết quả nhé!

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

<=> \(\left(\sin x-\cos x\right)\left(1+\cos x.\sin x\right)+\left(\cos x-\sin x\right)\left(\cos x+\sin x\right)=0\)

<=> \(\left(\sin x-\cos x\right)\left(\cos x+1\right)\left(\sin x+1\right)=0\)

2) \(\left(\sin^3x-2\sin^5x\right)-\left(2\cos^5x-\cos^3x\right)=0\)

<=> \(\sin^3x\left(1-2\sin^2x\right)-\cos^3x\left(2\cos^2x-1\right)=0\)

<=> \(\sin^3x.\cos2x-\cos^3x.\cos2x=0\)

<=> \(\cos2x\left(\sin^3x-\cos^3x\right)=0\)

3) ĐK: x\(\ne\frac{\pi}{2}+k\pi\)

\(\cos x\left(3.\tan x+2\right)-\left(3\tan x+2\right)=0\)

<=> \(\left(\cos x-1\right)\left(3.\tan x+2\right)=0\)

c/

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\frac{k\pi}{4}\)

\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\)

\(\Leftrightarrow4sinx.cos2x+2cos2x=2\)

\(\Leftrightarrow cos2x\left(2sinx+1\right)=1\)

\(\Leftrightarrow\left(1-2sin^2x\right)\left(2sinx+1\right)=1\)

\(\Leftrightarrow4sin^3x+2sin^2x-2sinx=0\)

\(\Leftrightarrow2sinx\left(2sin^2x+sinx-1\right)=0\)

\(\Leftrightarrow2sin^2x+sinx-1=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b/

Nhận thấy \(sinx=0\) không phải nghiệm, pt tương đương:

\(sinx.cosx.cos2x.cos4x.cos8x=\frac{1}{16}sinx\)

\(\Leftrightarrow8sin2x.cos2x.cos4x.cos8x=sinx\)

\(\Leftrightarrow4sin4x.cos4x.cos8x=sinx\)

\(\Leftrightarrow2sin8x.cos8x=sinx\)

\(\Leftrightarrow sin16x=sinx\)

\(\Rightarrow\left[{}\begin{matrix}16x=x+k2\pi\\16x=\pi-x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{15}\\x=\frac{\pi}{17}+\frac{k2\pi}{17}\end{matrix}\right.\)

a.

\(1-sin^2x+1-2sin^2x+sinx+2=0\)

\(\Leftrightarrow-3sin^2x+sinx+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{4}{3}\left(l\right)\end{matrix}\right.\)

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

b. ĐKXĐ; ...

\(5tanx-\frac{2}{tanx}-3=0\)

\(\Leftrightarrow5tan^2x-3tanx-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\frac{2}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-\frac{2}{5}\right)+k\pi\end{matrix}\right.\)

e.

Ko rõ vế phải

f.

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow1-2sin^22x=0\)

\(\Leftrightarrow cos4x=0\)

\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)