Hello,
For $\displaystyle \Theta $ $\displaystyle \epsilon $ [-2pi, 2pi] find all solutions of the trigonometric equation 2sin2 $\displaystyle \Theta $ - tan $\displaystyle \Theta $ = 0.
$\displaystyle
2 sin 2 \theta - tan \theta = 0
$
= $\displaystyle
4 sin \theta cos \theta - \frac {sin \theta}{cos \theta} = 0
$
= $\displaystyle
(sin \theta ) \frac{ 4 cos^2 \theta - 1 }{cos \theta} = 0
$
= $\displaystyle
\frac { (sin \theta )(2 cos \theta - 1 )(2 cos \theta + 1 ) }{cos \theta} = 0
$
= $\displaystyle
\frac { 4(sin \theta )( cos \theta - \frac{1}{2} )( cos \theta + \frac{1}{2} ) }{cos \theta} = 0
$
Check where the numerator is 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, $\displaystyle \sin \Theta$, equals zero. The rest follow similar logic.