A. Sin3x + Cos2x - Sinx = 0 B. Sin5x + 2cos^2x = 1 C. Sinx + 4cosx = 2 ...

Đáp án:

a)$\left[ \begin{array}{l}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7}{6}\pi+k2\pi\end{array} \right.$

b)$ \left[ \begin{array}{l}x=-\dfrac{\pi}{6}-\dfrac{k2\pi}{3}\\x=-\dfrac{\pi}{14}-\dfrac{k2\pi}{7}\end{array} \right.$

c)$\left[ \begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{array} \right.$

d)$\left[ \begin{array}{l}x=\dfrac{2\pi}{3}+k2\pi\\x=k2\pi\\x=\pi+k2\pi\end{array} \right. $

e)$ x=\dfrac{3\pi}{8}+k\pi$

f)$ x=\dfrac{\pi}{2}+k\pi$

Giải thích các bước giải:

a) $sin 3x+cos 2x-sin x=0$

$\Rightarrow (sin 3x-sin x)+cos 2x=0$

$\Rightarrow 2cos2x.sinx +cos2x=0$

$\Rightarrow cos 2x.(2sin x+1)=0$

$\Rightarrow \left[ \begin{array}{l}cos 2x=0\\2sin x+1=0\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}cos 2x=0\\sin x=-\dfrac{1}{2}\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}2x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7}{6}\pi+k2\pi\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7}{6}\pi+k2\pi\end{array} \right.$

b) $sin 5x+2cos^2x=1$

$\Rightarrow sin5x+2cos^2x-1=0$

$\Rightarrow sin5x+cos2x=0$

$\Rightarrow -sin5x=cos2x$

$\Rightarrow sin(-5x)=sin(\dfrac{\pi}{2}-2x)$

$\Rightarrow \left[ \begin{array}{l}-5x=\dfrac{\pi}{2}-2x+k2\pi\\-5x=\pi-\dfrac{\pi}{2}+2x+k2\pi\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}x=-\dfrac{\pi}{6}-\dfrac{k2\pi}{3}\\x=-\dfrac{\pi}{14}-\dfrac{k2\pi}{7}\end{array} \right.$

c) $sinx+4cosx=2+sin2x$

$\Rightarrow sinx+4cosx-2-2sinx.cosx=0$

$\Rightarrow sinx(1-2cosx)+2(2cosx-1)=0$

$\Rightarrow (2cosx-1)(2-sinx)=0$

$\Rightarrow \left[ \begin{array}{l}2cosx-1=0\\2-sinx=0\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}cosx=\dfrac{1}{2}\\sinx=2 (L)\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{array} \right.$

d) $cos3x +cos2x-cosx-1=0$

$4cos^3x-3cosx+2cos^2x-1-cosx-1=0$$\Rightarrow 4cos^3x+2cos^2x-4cosx-2=0$

$\Rightarrow 2cos^3x+cos^2x-2cosx-1=0$

$\Rightarrow cos^2x(2cosx+1)-(2cosx+1)=0$

$\Rightarrow (2cosx+1)(cos^2x-1)=0$

$\Rightarrow (2cosx+1)(cosx-1)(cosx+1)=0$

$\Rightarrow \left[ \begin{array}{l}2cosx+1=0\\cosx-1=0\\cosx+1=0\end{array} \right. $

$\Rightarrow \left[ \begin{array}{l}x=\dfrac{2\pi}{3}+k2\pi\\x=k2\pi\\x=\pi+k2\pi\end{array} \right. $

e) $sin^2(x-\dfrac{\pi}{4})=cos^2x$

$\Rightarrow \left[ \begin{array}{l}sin(x-\dfrac{\pi}{4})=cosx\\sin(x-\dfrac{\pi}{4})=-cosx\end{array} \right.$

$\Rightarrow \left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{\pi}{2}-x+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{\pi}{2}+x+k2\pi\end{array} \right.$

$\Rightarrow x=\dfrac{3\pi}{8}+k\pi$

f) $5cosx-sin2x=0$

$\Rightarrow 5cosx-2sinx.cosx=0$

$\Rightarrow cosx(5-2sinx)=0$

$\Rightarrow \left[ \begin{array}{l}cosx=0\\sinx=\dfrac{5}{2}(L)\end{array} \right.$

$\Rightarrow x=\dfrac{\pi}{2}+k\pi$