Question

Solve the following equations.
$\sqrt{3}\cos 7x+sin9x=sin7x+\sqrt{3}cos9x.$

Answer 1

Question $sin9x+\sqrt{3}cos7x=sin7x+\sqrt{3}cosx$

sol : Squaring both sides,

$(sin9x+\sqrt{3}cos7x{)}^2=(sin7x+\sqrt{3}cos9x{)}^2$

$\Rightarrow si{n}^{2}9x+3co{s}^{2}7x+2\sqrt{3}cos7xsin9x=si{n}^{2}2x+3cos9x+2\sqrt{3}sin2xcos9x$

$\Rightarrow si{n}^{2}3x+3co{s}^{2}7x-si{n}^{2}7x-3co{s}^{2}9x+3-3=-2\sqrt{3}(cos7xsin9x-sin7xcos9x)$

$\Rightarrow 4si{n}^{2}9x-4si{n}^{2}7x=2\sqrt{3}sin(-2x)$

$\Rightarrow 4(sin2x+sin7x)(sin9x-sin7x)=2\sqrt{3}sin2x)$

$\Rightarrow 16sin8xcosxsinxcos8x=-2\sqrt{3}sin2x$

$\Rightarrow 16sin8xcos8xcosxsinx=-4\sqrt{3}sinxcosx$

$\Rightarrow 4sin8xcos8x=-\sqrt{3}$

$\Rightarrow 2sin\left(16x\right)=-\sqrt{3}$

$\Rightarrow sin\left(16x\right)=\frac{-\sqrt{3}}{2}$

General solution for $sin\theta =sinx$

$\theta =n\pi \pm \left(x\right)$

$\therefore 16x=n\pi \pm (-\frac{\pi }{3})$

$x=\frac{n\pi }{16}\ne \frac{\pi }{48}$