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I am not sure how to solve this: $$\sin^2 (2x) = 2 \sin (2x)$$ I thought I could rewrite it like this: $$4 \sin^2 (x) \cos^2 (x) = 4 \sin (x) \cos (x)$$ and maybe like this as well: $$ \sin (x) \cos (x) = 0$$

but I have no idea whether it is correct and how get to the solution ..

thanks for help

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  • $\begingroup$ hint: if $a^2=2a$ then $a=2$ (if possible) or $a=0$. Apply this on $a=\sin2x$ $\endgroup$
    – drhab
    Commented Oct 15, 2018 at 14:28

3 Answers 3

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HINT:

Let $\sin{(2x)}=t$ and you get a quadratic equation in $t$.

You need to calculate roots of, $t^2-2t=0$.

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$$\sin^2(2x) = 2\sin(2x)$$ Assign a variable to $\sin(2x)$ to get a simple quadratic equation. $$t = \sin(2x)$$ Rewrite the equation. $$t^2 = 2t$$ Now, just solve the equation. $$t^2-2t = 0 \implies t(t-2) = 0 \implies t = 0 \text{ OR } t = 2$$ Plug in $\sin(2x)$. For $n \in \mathbb{Z}$, we get the following solutions: $$t = 0 \implies \sin(2x) = 0 \implies 2x = (\sin^{-1} 0)+2\pi n$$

$$\implies \begin{cases} 2x = 2\pi n \implies \boxed{x = \pi n}\\ \\ 2x = \pi+2\pi n\implies \boxed{x = \frac{\pi}{2}+\pi n}\\ \end{cases}$$ We can immediately rule out the second option ($t = 2$) because the range of $\sin x$ is $y \in [-1, 1]$.

Combining the two solutions, we can reach a single general solution: $$\boxed{x = \frac{n\pi}{2}}$$

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  • $\begingroup$ You could also immediately say that $2x = \pi n \implies x = \frac{\pi n}{2}$, which is just better. $\endgroup$
    – KM101
    Commented Oct 15, 2018 at 15:18
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HINT

We have

$$\sin^2 (2x) = 2 \sin (2x)\iff \sin^2 (2x) - 2 \sin (2x)=0 \iff \sin (2x)\cdot [\sin (2x) - 2]=0$$

then recall that

$$A\cdot B=0 \iff A=0 \,\lor\,B=0$$

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