Chapter EP, Problem 14.5.1EP

To determine

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

Answer to Problem 14.5.1EP

The solution of the equation is $\text{θ=}\frac{\text{5π}}{\text{6}}$ and $\text{θ=}\frac{\text{7π}}{\text{6}}$ .

Explanation of Solution

Given:

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

Concept used:

  $\text{cosθ=x}$ .

Which implies:

  ${\text{θ=cos}}^{-1}\text{x}$ .

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

Calculation:

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

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

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

The extract value of ${\text{cos}}^{\text{-1}}\left(\text{-}\frac{\sqrt{\text{3}}}{\text{2}}\right)$ is $\frac{\text{5π}}{\text{6}}$ .

  $\text{θ=}\frac{\text{5π}}{\text{6}}$ .

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

  $\text{θ=2π-}\frac{\text{5π}}{\text{6}}$ .

  $\text{θ=}\frac{\text{7π}}{\text{6}}$ .

Finding the period of the $\text{cosθ}$ which is $\text{2πrad}$ in both directions.

$\text{θ=}\frac{\text{5π}}{\text{6}}\text{+2πn,}\frac{\text{7π}}{\text{6}}\text{+2πn}$ for any integer $\text{n}$ .

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

Hence, the solution of the equations is $\text{θ=}\frac{\text{5π}}{\text{6}}$ and $\text{θ=}\frac{\text{7π}}{\text{6}}$ .