Find the general solution of the equation sin | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\sin x+\sin 3 x+\sin 5 x=0$

$(\sin x+\sin 5 x)+\sin 3 x=0$

$\Rightarrow\left[2 \sin \left(\frac{x+5 x}{2}\right) \cos \left(\frac{x-5 x}{2}\rig

Vedclass · Accepted Answer

Vedclass · Answer

$\sin x+\sin 3 x+\sin 5 x=0$

$(\sin x+\sin 5 x)+\sin 3 x=0$

$\Rightarrow\left[2 \sin \left(\frac{x+5 x}{2}\right) \cos \left(\frac{x-5 x}{2}\right)\right]+\sin 3 x=0$ $\left[\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$

$\Rightarrow 2 \sin 3 x \cos (-2 x)+\sin 3 x=0$

$\Rightarrow 2 \sin 3 x \cos 2 x+\sin 3 x=0$

$\Rightarrow \sin 3 x(2 \cos 2 x+1)=0$

$\Rightarrow \sin 3 x=0 \quad$ or $\quad 2 \cos 2 x+1=0$

Now, $\sin 3 x=0 \Rightarrow 3 x=n \pi,$ where $n \in Z$

i.e., $x=\frac{n \pi}{3},$ where $n \in Z$

$2 \cos 2 x+1=0$

$\Rightarrow \cos 2 x=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)$

$\Rightarrow \cos 2 x=\cos \frac{2 \pi}{3}$

$\Rightarrow 2 x=2 n \pi \pm \frac{2 \pi}{3},$ where $n \in Z$

$\Rightarrow x=n \pi \pm \frac{\pi}{3},$ where $n \in Z$

Therefore, the general solution is $\frac{n \pi}{3}$ or $n \pi \pm \frac{\pi}{3}, n \in Z$

Find the general solution of the equation $\sin x+\sin 3 x+\sin 5 x=0$

Similar Questions

Let $\theta, 0 < \theta < \pi / 2$, be an angle such that the equation $x ^2+4 x \cos \theta+\cot \theta=0$ has equal roots for $x$. Then $\theta$ in radians is

$\cot \theta = \sin 2\theta (\theta \ne n\pi $, $n$ is integer), if $\theta = $

The number of solutions that the equation $sin5\theta cos3\theta = sin9\theta cos7\theta $ has in $\left[ {0,\frac{\pi }{4}} \right]$ is

The number of real solutions $x$ of the equation $\cos ^2(x \sin (2 x))+\frac{1}{1+x^2}=\cos ^2 x+\sec ^2 x$ is

Number of solution$(s)$ of the equation $ln(1 + sin^2x) = 1 -ln(5 + x^2)$ is -