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

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

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

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

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

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

\(\Leftrightarrow sin^6x-sin^{10}x+cos^6x-cos^{10}x=0\)

\(\Leftrightarrow sin^6x\left(1-sin^4x\right)+cos^6x\left(1-cos^4x\right)=0\)

Do \(\left\{{}\begin{matrix}1-sin^4x\ge0\\1-cos^4x\ge0\end{matrix}\right.\) \(\forall x\) nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin^6x\left(1-sin^4x\right)=0\\cos^6x\left(1-cos^4x\right)=0\end{matrix}\right.\)

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

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

1.

\(\Leftrightarrow3x=k\pi\Leftrightarrow x=\frac{k\pi}{3}\)

2.

\(\Leftrightarrow cos5x=0\Leftrightarrow5x=\frac{\pi}{2}+k\pi\Leftrightarrow x=\frac{\pi}{10}+\frac{k\pi}{5}\)

4.

\(cos3x+cosx+cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos2x=0\)

\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

5.

\(sin6x+sin2x+sin4x=0\)

\(\Leftrightarrow2sin4x.cos2x+sin4x=0\)

\(\Leftrightarrow sin4x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\)

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

6. ĐKXĐ; ...

\(\Leftrightarrow tanx+tan2x=1-tanx.tan2x\)

\(\Leftrightarrow\frac{tanx+tan2x}{1-tanx.tan2x}=1\)

\(\Leftrightarrow tan3x=1\)

\(\Leftrightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)

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:

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

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

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

\(\Leftrightarrow2sin8x.cos8x=sinx\)

\(\Leftrightarrow sin16x=sinx\)

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

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

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\(\ne\frac{\pi}{2}+k\pi\)

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

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

1.

Các hàm \(sinx;sin\frac{x}{2};sin\frac{x}{3};...;sin\frac{x}{10}\) có chu kì lần lượt là \(2\pi;4\pi;6\pi;...;20\pi\)

\(\Rightarrow\) Chu kì của hàm đã cho là \(BCNN\left(2\pi;4\pi;...;20\pi\right)=15120\pi\)

2.

a.

\(y=cos^22x+3cos2x+3\)

\(y=\left(cos2x+1\right)\left(cos2x+2\right)+1\ge1\Rightarrow y_{min}=1\) khi \(cos2x=-1\)

\(y=\left(cos2x-1\right)\left(cos2x+4\right)+7\le7\Rightarrow y_{max}=7\) khi \(cos2x=1\)

b.

Đặt \(a=4sinx-3cosx\Rightarrow a^2\le\left(4^2+\left(-3\right)^2\right)\left(sin^2x+cos^2x\right)=25\)

\(\Rightarrow-5\le a\le5\)

\(y=a^2-4a+1\) với \(a\in\left[-5;5\right]\)

\(y=\left(a-2\right)^2-3\ge-3\Rightarrow y_{min}=-3\) khi \(a=2\)

\(y=\left(a-9\right)\left(a+5\right)+46\le46\Rightarrow y_{max}=46\) khi \(a=-5\)