Chapter EP, Problem 14.5.2EP

To determine

To find: the all solution of the equation given below.

Answer to Problem 14.5.2EP

The solution of the equations is $\text{θ=-}\frac{\text{π}}{6}$ and $\text{θ=}\frac{\text{5π}}{3}$ .

Explanation of Solution

Given:

  $\text{sin2θ=-}\frac{\sqrt{\text{3}}}{\text{2}}$ .

Concept used:

If $\text{sin2θ=x}$ then ${\text{2θ=sin}}^{-1}\text{x}$ .

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the from $\text{2π}$ to find the reference angle. next, add this reference angle to $\text{π}$ just to find the solution in third quadrant.

Calculation:

  $\text{sin2θ=-}\frac{\sqrt{\text{3}}}{\text{2}}$ .

Take the inverse sine of both sides of the equation to extract $\text{θ}$ from inside the sine.

  ${\text{2θ=sin}}^{\text{-1}}\left(\text{-}\frac{\sqrt{\text{3}}}{\text{2}}\right)$ .

The extract value of ${\sin }^{\text{-1}}\left(\text{-}\frac{\sqrt{\text{3}}}{\text{2}}\right)$ is $-\frac{\text{π}}{3}$ .

  $\text{2θ=-}\frac{\text{π}}{3}$ .

Dividing each term by $\text{2}$ and simplify:

  $\text{θ=-}\frac{\text{π}}{6}$ .

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the from $\text{2π}$ to find the reference angle. next, add this reference angle to $\text{π}$ just to find the solution in third quadrant.

  $\Rightarrow \text{2θ=2π+}\frac{\text{π}}{\text{3}}\text{+π}$ .

  $\Rightarrow \text{θ=}\frac{\text{5π}}{3}$ .

  $\text{θ=}\frac{\text{5π}}{3}$

Since, ${\text{θ lies in 0}}^{\text{0}}{\text{<θ<360}}^{\text{0}}$ .

Hence, the solution of the equations is $\text{θ=-}\frac{\text{π}}{6}$ and $\text{θ=}\frac{\text{5π}}{3}$ .