Trigonometric equation $2\sin(2x)+3=2\sin^2(x)$

Question

How to solve the following trigonometric equation:

$$2\sin(2x)+3=2\sin^2(x)$$

I tried:

1. Got everything on one side
2. divided by $2$ and got: $-\sin^2x+\sin2x+3/2=0$
3. I used double angle formulas to: $-\sin^2x+2\sin x\cos x+3/2=0$

4.How to continue?

Answer 1

Substitute $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$ $$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$ and Substitute $$\tan(\frac{x}{2})=t$$ and solve the equation in $t$

Answer 2

Replace the lone $3$ with $3\cos^2 x + 3\sin^2 x$. Rearrange:

$$\sin^2 x + 4\sin x \cos x + 3\cos^2 x = 0.$$

Factor:

$$(\sin x +\cos x)(\sin x + 3\cos x) = 0.$$

So either $\sin x = -\cos x$ and $x = 3\pi/4$ (and its relatives) or

$\sin x = -3 \cos x$ and $x = \arctan( -3)$ (and its relatives.)

Answer 3

Just in case you are interested in yet another method, you can write the equation as $$2\sin 2x+3=1-\cos 2x$$ $$\implies 2\sin 2x+\cos 2x=-2$$

Then use a compound angle transformation to get $$\sqrt{5}\cos(2x-\alpha)=-2$$ with $\tan\alpha=2$ and go ahead and solve.