How to go about solving $\sin^2(2x) - \sin^2(x) = \frac{1}{4}$

Question

Should I try to factor this?

$$ \sin^2(2x) - \sin^2(x) = \frac{1}{4} $$

Answer 1

Hint: use the double-angle formula and express the left side in terms of $\sin(x)$.

Answer 2

From $\color{red}{\sin^2(2x)=1-\cos^2(2x)}$ and $\color{red}{\sin^2x=\dfrac{1-\cos(2x)}{2}}$ we write $$1-\cos^2(2x)-\dfrac{1-\cos(2x)}{2}=\dfrac14$$ with $\color{blue}{\cos(2x)=u}$ say $$1-u^2-\dfrac{1-u}{2}=\dfrac14$$ or $$-4u^2+2u+1=0$$ after solving $$\color{blue}{u=\frac{2-\sqrt{5}}{4}}~~~,~~~u=\frac{2+\sqrt{5}}{4}>1$$ then $$\cos(2x)=\frac{2-\sqrt{5}}{4}$$ and $$x=\frac12\arccos\frac{2-\sqrt{5}}{4}$$

Answer 3

Partial solution:

We have $\sin^2(2x)=4\sin^2(x)\cos^2(x)=4\sin^2(x)(1-\sin^2(x)).$

Let $\sin^2(x)=y.$ then $4y(1-y)-y=\frac{1}{4}$. So $4y^2-3y+\frac{1}{4}=0.$

Using the quadratic formula on $16y^2-12y+1=0$ gives $y=\frac{12\pm\sqrt{144-64}}{32}=\frac{12\pm4\sqrt{5}}{32}=\frac{3\pm\sqrt{5}}{8}.$

From here it's easy.

Answer 4

Transform the full expression into a polynomial with respect to $\sin(x)$.

$$\sin^2(2x)-\sin^2(x)=\frac{1}{4}$$

Use the identity $\sin(2x)\equiv 2\cos(x)\sin(x)$

$$4\cos^2(x)\sin^2(x)-\sin^2(x)=\frac{1}{4}$$

Now, use the identity $\sin^2(x)+\cos^2(x)\equiv 1$

$$4(1-\sin^2(x))\sin^2(x)-\sin^2(x)=\frac{1}{4}$$

$$3\sin^2(x)-4\sin^4(x)=\frac{1}{4}$$

Now, you can perform a substitution $u=\sin^2(x)$

$$3u-4u^2=\frac{1}{4}$$

Solve the quadratic.