1. Need help! (Precalculus)

I have 7 problems that I need help with. If you know just one, it would be EXTREMELY helpful!

1. Find all solutions to csc x= -1.15 in the interval 0<x<2pi.

2. Solve 15cos^2(x) - 7cosx-4=0 by factoring and find solutions in 0<x<2pi.

3. Use appropriate addition formula to simplify cos(A-B) given sec(A) = -2/√3 where 90°<A<180° and tan(B) = 1/4 where 0°<B<90°

4. Find all solutions of 2sin^2(x) - 9sin(x)+4=0. Express in radians

5. Find all solutions to sin(3x) = -√2/(2) in interval 0<x<2pi.

6. Find all solutions for 2sin(x) = √2cos(2x) in interval 0<x<2pi.

7. Verify the following identity: cos(4x) = 8cos^4(x) - 8cos^2(x) + 1.

Thank you!

Here's some help . . .

2. Solve $15\cos^2\!x - 7\cos x-4\:=\:0$ by factoring and find solutions in $0\,<\,x\,<\,2\pi$
They told you what to do ... where is your difficulty?

Factor: . $(5\cos x - 4)(3\cos x + 1) \:=\:0$

. . $5\cos x-4\;=\:0\quad\Rightarrow\quad\cos x\:=\:\frac{4}{5}\quad\Rightarrow\quad x\:\approx\:0.6435,\;5.6397$

. . $3\cos x+1\:=\:0\quad\Rightarrow\quad\cos x\:=\:-\frac{1}{3}\quad\Rightarrow\quad x\:\approx\:1.9106,\;4.3726$

7. Verify the following identity: . $\cos4x \:= \:8\cos^4\!x - 8\cos^2\!x + 1.$
Recall that: . $\cos2\theta \:=\:2\cos^2\!\theta-1$

We have: . $\cos4x \;=\;2\cos^2\!2x - 1$

. . . . . . . . . . . . $= \;2(2\cos^2\!x - 1)^2 - 1$

. . . . . . . . . . . . $= \;2(4\cos^4\!x - 4\cos^2\!x + 1) - 1$

. . . . . . . . . . . . $= \;8\cos^4\!x - 8\cos^2\!x + 2 - 1$

. . . . . . . . . . . . $= \;8\cos^4\!x - 8\cos^2\!x + 1$