Question

Solve the following equations.
Find the roots of the equation sin $(2x+\frac{\pi }{18})cos(2x-\frac{\pi }{9})=-\frac{1}{4}intheinterval(0,\frac{\pi }{2}).$

Answer 1

$sin(2x+\pi /18)cos(2x-\pi /9)=-1/4$

$\Rightarrow 2sin(2x+\pi /18)cos(2x-x/9)=-1/2$

$\Rightarrow sin(4x-\pi /18)+sin(\frac{\pi }{18}+\frac{\pi }{9})=\frac{-1}{2}$

(by $sinAcosB$ formula)

$\Rightarrow sin(4x-\frac{\pi }{18})+sin\left(\frac{3\pi }{18}\right)=-\frac{1}{2}$

$\Rightarrow sin(4x-\frac{\pi }{18})+\frac{1}{2}=-\frac{1}{2}$

$\Rightarrow sin(4x-\pi /18)=-1$

$\Rightarrow 4x-\pi /18=(2x+1)\pi /2$

$\Rightarrow 4x=(2x+1)\frac{\pi }{2}+\frac{\pi }{18}$

$\Rightarrow x=(2x+1)\frac{\pi }{8}+\frac{\pi }{72}$

if $xϵ(0,\pi /2)$

$x=0\Rightarrow x=\pi /8+\pi /72=10\pi /72$

$x=1\Rightarrow x=\frac{3\pi }{8}+\frac{\pi }{72}=28\pi /72$

$x=2\Rightarrow x=\frac{5\pi }{8}+\frac{\pi }{72}=46\pi /72$

$x=3\Rightarrow x=\frac{7\pi }{8}+\frac{\pi }{72}=64\pi /72$