Question

Solve the following equations.
$cos(2sinx+(1+\sqrt{3}\left)cosx\right)=sin\left(\right(1-\sqrt{3}\left)cosx\right).$

Answer 1

$cos(2sinx+(1+\sqrt{3}\left)cosx\right)=sin(1-\sqrt{3}cosx)$

$\because sin(\frac{\pi }{2}+x)=cox$

we can write

$\therefore sin(\frac{\pi }{2}+2sinx+(1+\sqrt{3}\left)cosx\right)=sin\left(\right(1-\sqrt{3}\left)cosx\right)$

$\Rightarrow \frac{\pi }{2}+2sinx+(1+\sqrt{3})cosx=\left(\right(1-\sqrt{3}\left)cox\right)$

$\Rightarrow \frac{\pi }{2}+2sinx=(1-\sqrt{3})cosx-(1+\sqrt{3})cosx$

$\Rightarrow \frac{\pi }{2}+2sinx=-2\sqrt{3}cosx$

$\Rightarrow \frac{\pi }{2}+2sinx+2\sqrt{3}cosx=0$

$\Rightarrow 2(sinx+\sqrt{3}cosx)=-\pi /2$

$2\times 2(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx)=-\frac{\pi }{2}$

$\Rightarrow 4(cos\left(\frac{\pi }{3}\right)sinx+sin\frac{\pi }{3}cosx)=-\frac{\pi }{2}$

$\because sinacosb+cosasinb=sin(a+b)$

i.e

$\Rightarrow 4sin(\frac{\pi }{3}+x)=\frac{-\pi }{3}$

$\Rightarrow sin(\frac{\pi }{3}+x)=-\frac{\pi }{8}$

$\frac{\pi }{3}+x=si{n}^{-1}(-\pi /8)$

$x=-\frac{\pi }{3}-si{n}^{-1}(-\pi /8)$

$x=\frac{-\pi }{3}-si{n}^{-1}\left(\pi /8\right)$