# Thread: 5 Trig questions! (re-post)

1. ## 5 Trig questions! (re-post)

I need help with these following trig problems..

$2 - \tan \theta = 2 - \frac{\sin\theta}{\cos\theta} = \frac{2\cos\theta - \sin \theta}{\cos\theta}$

$2\csc \theta - \sec \theta = 2\frac1{\sin\theta} - \frac1{\cos\theta} = \frac{2\cos \theta - \sin \theta}{\sin\theta\cos\theta}$

Now substitute the above expressions:
$\frac{2 - \tan \theta}{2\csc \theta - \sec \theta} = \frac{\frac{2\cos\theta - \sin \theta}{\cos\theta}}{\frac{2\cos \theta - \sin \theta}{\sin\theta\cos\theta}} = \sin \theta$

Do as the hint says:

$2u^2 +5u + 2 = 0$
$\Rightarrow u = \frac{-5\pm\sqrt{5^2 - 4(2)(2)}}{2(2)} = \frac{-5\pm3}{4} = -\frac12,-2$

But $\cos x \geq -1$, -2 root can be discarded.

Thus you have to solve $\cos x = -\frac12$.

I am sure you can handle it from here.