Chapter 7, Problem 25CT

To determine

The solution of the equation $4{\sin }^2\theta -3=0$ on $0\le \theta <2\pi$.

Answer to Problem 25CT

Solution:

The solutions of equation $4{\sin }^2\theta -3=0$ on $0\le \theta <2\pi$ are $\theta =\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3}$ and $\frac{5\pi }{3}$.

Explanation of Solution

Given information:

The equation is $4{\sin }^2\theta -3=0$

Explanation:

Consider the equation $4{\sin }^2\theta -3=0$.

Adding $3$ from both sides,

$\Rightarrow 4{\sin }^2\theta =3$

Dividing both sides by $4$,

$\Rightarrow {\sin }^2\theta =\frac{3}{4}$

Taking the square root from both sides,

$\Rightarrow \sin \theta =\pm \frac{\sqrt{3}}{2}$

Since sine is positive in the first and second quadrant

$\sin \theta =\frac{\sqrt{3}}{2}$ implies that $\theta =\frac{\pi }{3},\frac{2\pi }{3}$ on $0\le \theta <2\pi$.

Since sine is negative in the third and fourth quadrant.

Hence, $\sin \theta =-\frac{\sqrt{3}}{2}$ implies that $\theta =\frac{4\pi }{3},\frac{5\pi }{3}$ on $0\le \theta <2\pi$.

Therefore, the solution of the equation $4{\sin }^2\theta -3=0$ on $0\le \theta <2\pi$ are $\theta =\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3}$ and $\frac{5\pi }{3}$.