Question

the no. of solutions of equation sin2x-2cosx+4sinx=4​

Answer 1

$x=n\pi+(-1)^n\dfrac{\pi}{2}$

n is natural number.

Step-by-step explanation:

Given: $\sin 2x-2\cos x+4\sin x=4$

But sin 2x = 2 sin x cos x

$2\sin x\cos x-2\cos x+4\sin x-4=0$

Factor the term

$2\cos x(\sin x-1)+4(\sin x-1)=0$

$(\sin x-1)(2\cos x+4)=0$

Equare each factor to 0

$\sin x-1=0\text{ or }2\cos x+4=0$

$\sin x=1$            $\cos x\neq-2$

$\text{ If }\sin \theta=\sin\alpha\text{ then }\theta=n\pi +(-1)^\alpha$

$x=n\pi+(-1)^n\dfrac{\pi}{2}$      

n is natural number

Exact solution: $x=\dfrac{\pi}{2}$

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