Question

Solve cos (x) tan (x) -1/2=0 over the interval [0,2π].

Answer 1

First, we are going to add $\frac{1}{2}$ from both sides of the equation: 
$cos(x)tan(x)- \frac{1}{2} + \frac{1}{2} = \frac{1}{2}$
$cos(x)tan(x)= \frac{1}{2}$

Next, we are going to use the trig identity: $tan(x)= \frac{sin(x)}{cos(x)}$ to rewrite our expression:
$cos(x)tan(x)= \frac{1}{2}$
$cos(x) \frac{sin(x)}{cos(x)} = \frac{1}{2}$
$sin(x)= \frac{1}{2}$

Finally, using our unitary circle, we can infer that $sin(x)= \frac{1}{2}$ from 0 to $2 \pi$ when $x= \frac{ \pi }{6}$ and $x= \frac{5 \pi }{6}$

We can conclude that the solutions of the equation cos (x) tan (x) -1/2=0 over the interval [0,2π] are: $x=\frac{ \pi }{6},\frac{5 \pi }{6}$