Solve on the interval $[0,2\pi)$: $4 \sin(x) \cos(x)=1$.

Question

I tried using the product-to-sum formulas, but did not come up with the correct answer.

Answer 1

As was hinted, $$\sin2x=2\sin x\cos x$$ Hence, your equation becomes $$2\sin2x=1$$ $$\sin2x=\frac12$$ $$2x=\arcsin\frac12$$ $$2x=\frac\pi6,\,\frac{5\pi}6$$ $$x=\frac\pi{12},\,\frac{5\pi}{12}$$

Answer 2

Hint: use the double-angle formula for sine:

$$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$