Question

Four times the sum of the roots of the equation $sin2x+5sinx+5cosx+1=0$ in the interval [0, 50$\pi$] is p$\pi$ where $p$ is equal to

Answer 1

Given, $\sin 2x+5\sin x+5\cos x+1=0$

We know that $si{n}^{2}x+{\cos }^{2}x=1$

Putting this in the above equation, we get

$si{n}^{2}x+{\cos }^{2}x+2\sin x\cos x+5\sin x+5\cos x=0(\sin x+\cos x{)}^2+5(\sin x+\cos x)=0(\sin x+\cos x\left)\right(\sin x+\cos x+5)=0$

So $\sin x=-\cos xand\sin x+\cos x=-5$

Second is not possible so we get

$\tan x=-1$

So Solutions in the interval $[0,50\pi ]$ are $[\pi +\pi /4,2\pi -\pi /4,3\pi +\pi /\mathrm{4........},49\pi +\pi /4,50\pi -\pi /4]$

So the Sum of all the solutions is

$S=\pi \left(50\right)\left(51\right)/24S=5100\pi =p\pi$

So$p=5100$