Question

Solve the following equations.
$\sqrt{3}sin2x+cos5x-cos9x=0.$

Answer 1

$\sqrt{3}sin2x+(cos5s-cos9x)=0$

$\{\because cosC-cosD=2sin\left(\frac{C+D}{2}\right).sin\left(\frac{D-C}{2}\right)\}$

$\Rightarrow \sqrt{3}sin2x+2sin\left(\frac{5x+9x}{2}\right).sin\left(\frac{9x-5x}{2}\right)=0$

$\Rightarrow \sqrt{3}sin2x+2sin7x.sin2x=0.$

$\Rightarrow sin2x(\sqrt{3}+2sin7x)=0$

$\Rightarrow sin2x=0$ or $\sqrt{3}+2sin7x=0$

$\Rightarrow sin2x=0$ or $sin7x=\frac{-\sqrt{3}}{2}$

we know, the general solution when

$sin\theta =0\Rightarrow \{\theta =n\pi ,nϵZ\},Z\to$ integer $(0.\pm 1,\pm 2,\pm 3\pm ...)$

For $\theta =2x$

$sin2x=0\Rightarrow \{2x=n\pi ,nϵZ\}$

$\Rightarrow \{x=n\frac{\pi }{2},nϵZ\}$

also we know general solution when

$sin\theta =sin\alpha =\{\theta =n\pi +(-1{)}^n\alpha ,nϵZ\}...\left(1\right)$

For $\theta =7x\alpha =4\pi /3$

$\because utsiny=-\sqrt{3}/2$

$\because$ it is must be in III and IV quadrant

$\Rightarrow$ also $sin\frac{\pi }{3}=\frac{\sqrt{3}}{2}$

$\Rightarrow siny=\frac{-\sqrt{3}}{2}\Rightarrow siny=sin(\pi +\frac{\pi }{3})=sin4\frac{\pi }{3}$

$\Rightarrow sin7x=sin\left(4\pi /3\right)$

Here $\theta =7x$ and $\alpha =4\pi /3$

using (1) general solution of $sin7x=sin\left(4\pi /3\right)$ will be

$7x=n\pi +(-1{)}^n\frac{4\pi }{3},nϵZ$

$\Rightarrow x=n\frac{\pi }{7}+(-1{)}^{n}4\frac{\pi }{21},nZ$