Question

How do you find all solutions between 0 to $2\pi$ for $4{\sin }^{2}x-3=0$ ?

Answer 1

Hint: In the equation $4{\sin }^{2}x-3=0$ first we will find the value of sin x and then we will check that value whether that value lie in the range of sin x , we know that range of sin x is from -1 to 1 . If the roots come in the range of sin x then we will solve for x by using graphs.

Complete step by step answer:
The given equation in the question is $4{\sin }^{2}x-3=0$ we have found all values of x that satisfy the equation and are between 0 to $2\pi$ .
First let’s find the value of sin x that satisfy the equation
Let’s add 3 in both LHS and RHS , by adding 3 both sides we get
$4{\sin }^{2}x=3$
Diving LHS and RHS by 4 we get
${\sin }^{2}x={\displaystyle \frac{3}{4}}$
So the value of sin x is $\pm {\displaystyle \frac{\sqrt{3}}{2}}$
So we have find the value of x for which sin x is equal to ${\displaystyle \frac{\sqrt{3}}{2}}$ and sin x is $-{\displaystyle \frac{\sqrt{3}}{2}}$
Let’s solve this by graph

We can see the point of intersection the 2 line with y= sin x are A , B, C, and D
So all the solutions from 0 to $2\pi$ are ${\displaystyle \frac{\pi }{3}}$ , ${\displaystyle \frac{2\pi }{3}}$ , ${\displaystyle \frac{4\pi }{3}}$ and ${\displaystyle \frac{5\pi }{3}}$

Note:
We can check all the solutions whether these are correct or not by putting the value of x in the equation. When we put ${\displaystyle \frac{\pi }{3}}$ or ${\displaystyle \frac{2\pi }{3}}$ in the equation $4{\sin }^{2}x-3=0$ the result came out to be $4{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2-3$ which is equal to 0 , so ${\displaystyle \frac{\pi }{3}}$ and ${\displaystyle \frac{2\pi }{3}}$ are correct answers. When we put ${\displaystyle \frac{4\pi }{3}}$ or ${\displaystyle \frac{5\pi }{3}}$ we get $4{(-{\displaystyle \frac{\sqrt{3}}{2}})}^2-3$ which is equal to 0, so ${\displaystyle \frac{4\pi }{3}}$ and ${\displaystyle \frac{5\pi }{3}}$ are correct answers.