Question

The number of solutions of the system of equations: $2\sin^2 x + \sin^2 2x = 2$ In [0, 4$\pi$] is ______.

Answer 1

12

Answer 2

We start with

$$2\sin^2x+\sin^2 2x=2.$$

Using the identity $\sin^2 2x = 4\sin^2x\cos^2x$, the equation becomes

$$2\sin^2x+4\sin^2x\cos^2x=2.$$

Dividing by 2:

$$\sin^2x+2\sin^2x\cos^2x=1.$$

Replace $\cos^2x$ with $1-\sin^2x$:

$$\sin^2x + 2\sin^2x(1-\sin^2x)=1.$$

Let $u=\sin^2x$. Then:

$$u+2u(1-u)=1 \quad\Rightarrow\quad u+2u-2u^2=1,$$ $$3u-2u^2=1 \quad\Rightarrow\quad 2u^2-3u+1=0.$$

Factorizing:

$$(2u-1)(u-1)=0.$$

Thus,

$$u=\frac{1}{2}\quad \text{or}\quad u=1.$$

Returning to $\sin^2x$:

• For $\sin^2x=\frac{1}{2}$: $\sin x=\pm \frac{1}{\sqrt{2}}$.
• For $\sin^2x=1$: $\sin x=\pm1$.

Counting solutions in $[0,4\pi]$:

• $\sin^2x=\frac{1}{2}$: In one period $[0,2\pi]$ there are 4 solutions: $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$. Over $[0,4\pi]$ there are $4 \times 2 = 8$ solutions.
• $\sin x=\pm1$: In $[0,2\pi]$ there are 2 solutions: $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. Over $[0,4\pi]$ there are $2 \times 2 = 4$ solutions.

Total solutions: $8+4=12$.

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Explanation (minimal): Substitute $\sin^2 2x=4\sin^2x\cos^2x$, express $\cos^2x$ in terms of $\sin^2x$, solve the quadratic $2u^2-3u+1=0$ to get $u=\frac{1}{2}$ and 1, then count solutions for $\sin^2x=\frac{1}{2}$ and $\sin^2x=1$ in $[0,4\pi]$.