Find all solutions of the equation in the interval [0, 2π]. 2sinx cos2x + 2cosx sin2x = -√2. Write your answer in radians in terms of π.

Steps to Solve

1. Rewrite the Equation Rewrite the original equation to simplify it. The equation is: $$ 2 \sin x \cos 2x + 2 \cos x \sin 2x = -\sqrt{2} $$

2. Factor Out Common Terms Notice that both terms on the left have a common factor of 2: $$ 2 (\sin x \cos 2x + \cos x \sin 2x) = -\sqrt{2} $$ Dividing both sides by 2 gives: $$ \sin x \cos 2x + \cos x \sin 2x = -\frac{\sqrt{2}}{2} $$

3. Use the Sine Addition Formula Recognize that $\sin x \cos 2x + \cos x \sin 2x$ can be rewritten using the sine addition formula: $$ \sin(x + 2x) = \sin(3x) $$

4. Set the New Equation Now we have: $$ \sin(3x) = -\frac{\sqrt{2}}{2} $$

5. Find General Solutions The solutions for $\sin \theta = -\frac{\sqrt{2}}{2}$ are: $$ 3x = \frac{7\pi}{4} + 2k\pi \quad \text{and} \quad 3x = \frac{5\pi}{4} + 2k\pi $$ for integer $k$.

6. Solve for x Divide the equations by 3: $$ x = \frac{7\pi}{12} + \frac{2k\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{12} + \frac{2k\pi}{3} $$

7. Find Solutions in the Interval [0, 2π] Substituting different integer values for $k$:

• For $k=0$:
  • $x = \frac{7\pi}{12}$
  • $x = \frac{5\pi}{12}$
• For $k=1$:
  • $x = \frac{7\pi}{12} + \frac{2\pi}{3} = \frac{7\pi}{12} + \frac{8\pi}{12} = \frac{15\pi}{12} = \frac{5\pi}{4}$
  • $x = \frac{5\pi}{12} + \frac{2\pi}{3} = \frac{5\pi}{12} + \frac{8\pi}{12} = \frac{13\pi}{12}$
• For $k=2$:
  • $x = \frac{7\pi}{12} + \frac{4\pi}{3} = \frac{7\pi}{12} + \frac{16\pi}{12} = \frac{23\pi}{12}$ (out of range)
  • $x = \frac{5\pi}{12} + \frac{4\pi}{3} = \frac{5\pi}{12} + \frac{16\pi}{12} = \frac{21\pi}{12}$ (out of range)

Thus, the valid solutions in the interval [0, 2π] are:

• $x = \frac{5\pi}{12}$
• $x = \frac{7\pi}{12}$
• $x = \frac{5\pi}{4}$
• $x = \frac{13\pi}{12}$