
5 Trig questions! (repost)

http://img522.imageshack.us/img522/3294/scan0002ll8.jpg
ANSWER:
$\displaystyle 2  \tan \theta = 2  \frac{\sin\theta}{\cos\theta} = \frac{2\cos\theta  \sin \theta}{\cos\theta}$
$\displaystyle 2\csc \theta  \sec \theta = 2\frac1{\sin\theta}  \frac1{\cos\theta} = \frac{2\cos \theta  \sin \theta}{\sin\theta\cos\theta}$
Now substitute the above expressions:
$\displaystyle \frac{2  \tan \theta}{2\csc \theta  \sec \theta} = \frac{\frac{2\cos\theta  \sin \theta}{\cos\theta}}{\frac{2\cos \theta  \sin \theta}{\sin\theta\cos\theta}} = \sin \theta$
http://img172.imageshack.us/img172/9089/scan0004ag6.jpg
Do as the hint says:
$\displaystyle 2u^2 +5u + 2 = 0$
$\displaystyle \Rightarrow u = \frac{5\pm\sqrt{5^2  4(2)(2)}}{2(2)} = \frac{5\pm3}{4} = \frac12,2$
But $\displaystyle \cos x \geq 1$, 2 root can be discarded.
Thus you have to solve $\displaystyle \cos x = \frac12$.
I am sure you can handle it from here.