1. ## [SOLVED] Trig Equation

Hello,

For $\Theta$ $\epsilon$ [-2pi, 2pi] find all solutions of the trigonometric equation 2sin2 $\Theta$ - tan $\Theta$ = 0.

2. $

2 sin 2 \theta - tan \theta = 0
$

= $

4 sin \theta cos \theta - \frac {sin \theta}{cos \theta} = 0
$

= $

(sin \theta ) \frac{ 4 cos^2 \theta - 1 }{cos \theta} = 0
$

= $

\frac { (sin \theta )(2 cos \theta - 1 )(2 cos \theta + 1 ) }{cos \theta} = 0
$

= $

\frac { 4(sin \theta )( cos \theta - \frac{1}{2} )( cos \theta + \frac{1}{2} ) }{cos \theta} = 0
$

Check where the numerator is 0.

3. Darn, I can't quite figure it out. Where does the numerator equal 0?

4. Originally Posted by OVechkin8
Darn, I can't quite figure it out. Where does the numerator equal 0?
If you have some number in the form 4ABC = 0, this will be true:

• if the first term, A, equals zero;
• if the second term, B, equals zero; OR
• if the third term, C, equals zero.

So above, at least one of your zeros will be at the place the first term, $\sin \Theta$, equals zero. The rest follow similar logic.

5. Originally Posted by OVechkin8
Darn, I can't quite figure it out. Where does the numerator equal 0?
where $\sin{\theta} = 0$ , $\cos{\theta} = \frac{1}{2}$ , and $\cos{\theta} = -\frac{1}{2}$