Question

Find values of  that satisfy the equation for 0º    360º and 0  θ  2 . Give answers in degrees and radians.

Answer 1

Answer:

$\theta = 30^\circ, 330^\circ$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

Step-by-step explanation:

Given [Missing from the question]

Equation:

$cos\theta = \frac{\sqrt 3}{2}$

Interval:

$0 \le \theta \le 360$

$0 \le \theta \le 2\pi$

Required

Determine the values of $\theta$

The given expression:

$cos\theta = \frac{\sqrt 3}{2}$

... shows that the value of $\theta$ is positive

The cosine of an angle has positive values in the first and the fourth quadrants.

So, we have:

$cos\theta = \frac{\sqrt 3}{2}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{\sqrt 3}{2})$

$\theta = 30$ --- In the first quadrant

In the fourth quadrant, the value is:

$\theta = 360 -30$

$\theta = 330$

So, the values of $\theta$ in degrees are:

$\theta = 30^\circ, 330^\circ$

Convert to radians (Multiply both angles by $\pi/180$)

So, we have:

$\theta = \frac{30 * \pi}{180}, \frac{330 * \pi}{180}$

$\theta = \frac{\pi}{6}, \frac{33 * \pi}{18}$

$\theta = \frac{\pi}{6}, \frac{11 * \pi}{6}$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$