What is the solution of the equation $\sin x \cos x -\sin^2 x =\frac14(\sqrt{6}-2)$?

Question

Frankly this question puzzled me too: How to solve the following equation :

$$\sin x \cos x -\sin^2 x =\frac{\sqrt{6}-2}{4}$$

I tried so much to no avail. I have reached this formula, but I did not know the solution

$$2\sin x+\cos x=\frac12 \sqrt{6}$$

Can you help me?

Answer 1

$$\sin(x)\cos(x) - \sin^2(x) = C$$

Where $C$ is your right hand side constant for simplicity.

Use trigonometric identities to rewrite it as

$$\frac{1}{2}\sin(2x) - \frac{1 - \cos(2x)}{2} = C$$

$$\sin(2x) + \cos(2x) = 2C + 1$$

Call $x = z/2$

$$\sin(z) + \cos(z) = 2C + 1$$

Square both

$$1 + 2\sin(z)\cos(z) = (2C +1)^2$$

$$\sin(2z) = (2C +1)^2 - 1$$

From here you can continue by yourself!

Answer 2

You need the duplication formulæ to solve: \begin{align} &\sin x\cos x - \sin^2x=\frac12\sin 2x-\frac{1-\cos 2x}2=\frac14(\sqrt 6-2)\\ \iff & \sin 2x+\cos 2x=\frac{\sqrt 6}2. \end{align}

Now $\;\sin 2x+\cos 2x=\sqrt 2\biggl(\dfrac{\sqrt 2}2\sin 2x+\dfrac{\sqrt 2}2\cos 2x\biggr)=\sqrt 2\sin\Bigl(2x+\dfrac\pi 4\Bigr)$, so the equation is equivalent to $$\sin\Bigl(2x+\dfrac\pi 4\Bigr)=\frac{\sqrt 3}2=\sin\frac\pi 3.$$