Question

2cos^2(3p/2-x)=-sinx

Answer 1

I hope this helps you

Answer 2

Let's tackle this equation step by step. First, we can simplify the left side using a trigonometric identity:

$2\cos^2\left(\frac{3\pi}{2} - x\right) = 1 + \cos\left(2\left(\frac{3\pi}{2} - x\right)\right)$

Now, let's substitute $1 + \cos\left(2\left(\frac{3\pi}{2} - x\right)\right)$ back into the original equation:

$1 + \cos\left(2\left(\frac{3\pi}{2} - x\right)\right) = -\sin x$

Now, isolate the cosine term:

$\cos\left(2\left(\frac{3\pi}{2} - x\right)\right) = -1 - \sin x$

Now, you can use the double-angle identity for cosine:

$\cos(2\theta) = 1 - 2\sin^2\theta$

Substitute $\frac{3\pi}{2} - x$ for $\theta$:

$1 - 2\sin^2\left(\frac{3\pi}{2} - x\right) = -1 - \sin x$

Simplify further:

$2\sin^2\left(\frac{3\pi}{2} - x\right) = 2 + \sin x$

Now, isolate the sine term:

$\sin^2\left(\frac{3\pi}{2} - x\right) = 1 + \frac{\sin x}{2}$

Now, take the square root of both sides:

$\sin\left(\frac{3\pi}{2} - x\right) = \pm\sqrt{1 + \frac{\sin x}{2}}$

Remember that $\sin\left(\frac{3\pi}{2} - x\right) = -\cos x$, so:

$-\cos x = \pm\sqrt{1 + \frac{\sin x}{2}}$

Now, you can square both sides to eliminate the square root and solve for $\sin x$. Keep in mind that the final solution may involve checking for extraneous solutions.