Solve sin(2x) cos(2x) = -1/2 for all real x.

Step-by-step solution

1. 1

   Recognise the product as half of a sine. A product $\sin\theta\cos\theta$ is never solved directly — it is always folded into a single sine using $\sin 2\theta = 2\sin\theta\cos\theta$. Here $\theta = 2x$, so $2\sin(2x)\cos(2x) = \sin(4x)$.

2. 2

   Multiply both sides by 2 to apply the identity.

   $2\sin(2x)\cos(2x) = 2\left(-\frac{1}{2}\right) \quad\Longrightarrow\quad \sin(4x) = -1$

   The equation now has one trig function of one argument, which is the form every general solution formula expects.

3. 3

   Solve sin(4x) = -1. The sine reaches its minimum $-1$ only at the bottom of each cycle, at angle $\tfrac{3\pi}{2}$ plus whole turns. Because $-1$ is an extreme value there is a single family, not the usual two:

   $4x = \frac{3\pi}{2} + 2\pi k, \qquad k \in \mathbb{Z}$

4. 4

   Divide by 4 to isolate x. Every term is divided, including the period:

   $x = \frac{3\pi}{8} + \frac{\pi}{2}k, \qquad k \in \mathbb{Z}$

   The period shrank from $2\pi$ to $\tfrac{\pi}{2}$, so on any interval of length $2\pi$ there are four solutions, not one.

5. 5

   Check the base solution. At $x = \tfrac{3\pi}{8}$ we get $2x = \tfrac{3\pi}{4}$, where $\sin\tfrac{3\pi}{4} = \tfrac{\sqrt2}{2}$ and $\cos\tfrac{3\pi}{4} = -\tfrac{\sqrt2}{2}$. Their product is $-\tfrac{2}{4} = -\tfrac12$, exactly the right-hand side.