Solving $4\sin^2\left(\frac\pi5(x-1)\right)=3$ with $x\in[0 , 2\pi)$

Question

$$4\sin^2\left(\frac\pi5(x-1)\right)=3\qquad x\in[0 , 2π)$$

My work:

$$\sin^2[\pi/5(x-1)] = 3/4$$

Let $p = \pi/5(x-1)$

$$\sin^2 p = 3/4$$

$$\sin p = \pm \sqrt{3}/2$$

$$p = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$$

$$\pi/5(x-1) = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$$

$$x = 8/3, 13/3, 23/3, 28/3$$

However, I'm not sure if these answers are correct or how many may be correct answers within the restricted range.

If, hypothetically, I wanted to set the range of $p$, what would it be without finding the correct $x$ values first?

Can you show how you would do the problem?

Answer 1

Since $x\in[0,2\pi)$, you must have $$ p=\frac{\pi}{5}(x-1)\in[a,b) $$ for some $a$ and $b$. Find $a$ and $b$ and then find $p$ such that $p\in[a,b)$ and $$ \sin p=\pm\frac{\sqrt{3}}{2}. $$ Once you have the values of $p$ in $[a,b)$, use the relation $$ p=\frac{\pi}{5}(x-1) $$ to the values of $x$.

Answer 2

For $\sin^2y=\sin^2A\iff\cos^2y=\cos^2A\iff\tan^2y=\tan^2A$

Method$\#1:$ use Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $

$\sin(y\pm A)=0,y=m\pi\mp A$

Method$\#2:$

Use $\cos2x=1-2\sin^2x,$

and $\cos2x=\cos2A\implies2x=2n\pi\pm2A$

Answer 3

When you get to $sin^2\left({\pi\over5}{(x-1)}\right)={3\over4}$,so using $sin^2(x)+cos^2(x)=1$,$cos\left({\pi\over5}{(x-1)}\right)={1\over2}$,$\left({\pi\over5}{(x-1)}\right)={\pi\over3}+2\pi n$. It is easy from here now.