1. ## solve

A.2sinx+1=0
B.2cos^2x-1=0
C.3tan^3x-tanx=0
D.2sin^2x=2+cosx

I really need help.. I have my answers here but im not sure if they are correct:

A.x=7/6pi, 11/6pi
B.x=pi/4 7/4pi
C.x=pi/6 7/6pi 2pi pi
D.x=pi/2, 3/2pi pi/3 5/3pi

I also need these checked
Using a sum of difference identity find exact values for the following:

A.Cos75(degrees)
B.Sin15(degrees)
C.tan345(degrees)

I got
A.*squareroot*6-2/4
B.*squareroot*6-2/4
C.3-*squareroot*3/3

For problems 2 and 3 find the following.
A.cos(x-y): sin(x-y); tan (x-y); the quadrant (x-y) is in.
B.cos(x+y); sin(x+y) tan(x+y) the quadrant (x+y) is in.

2.If (pi/2)<x<pi and pi<y<(3pi/2) and cosx=-5/13 and tany=3/4

3.If pi<x< or equal too(3pi/2) and (3pi/2)<y<2pi, and secx=-41/9 and tany=-12/5

If you could show me how to do them it would be appreciated.

2. Hello, ninelivesmeow!

I'll run through the first one . . .

$\displaystyle A)\;\;2\sin x+1\:=\:0$
We have: .$\displaystyle \sin x \:=\:-\frac{1}{2}\quad\Rightarrow\quad x \:=\:\frac{7\pi}{6},\:\frac{11\pi}{6}$

$\displaystyle B)\;\;2\cos^2\!x-1\:=\:0$
You missed a few . . .

We have: .$\displaystyle \cos^2\!x \:=\:\frac{1}{2}\quad\Rightarrow\quad \cos x \:=\:{\color{red}\pm}\frac{1}{\sqrt{2}}$

Therefore: .$\displaystyle \boxed{x \;=\;\frac{\pi}{4},\:\frac{3\pi}{4},\:\frac{5\pi}{ 4},\:\frac{7\pi}{4}}$

$\displaystyle C)\;\;3\tan^3\!x-\tan x\:=\:0$

We have: .$\displaystyle \tan x(3\tan^2\!x - 1) \:=\:0$

Then: .$\displaystyle \tan x \:=\:0\quad\Rightarrow\quad\boxed{ x \:=\:0,\:\pi,\:2\pi}$

And: .$\displaystyle \tan^2\!x \:=\: \frac{1}{3} \quad\Rightarrow\quad \tan x \:=\: {\color{red}\pm}\frac{1}{\sqrt{3}}\quad\Rightarrow \quad \boxed{x \;=\;\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{7\pi}{ 6},\:\frac{11\pi}{6}}$

$\displaystyle D)\;\;2\sin^2\!x\:=\:2+\cos x$
We have: .$\displaystyle 2(1-\cos^2\!x) \:=\:2+\cos x$

. . which simplifies to: .$\displaystyle 2\cos^2\!x + \cos x \:=\:0\quad\Rightarrow\quad \cos x(2\cos x + 1) \:=\:0$

Then: .$\displaystyle \cos x \:=\:0\quad\Rightarrow\quad \boxed{x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}$

And: .$\displaystyle 2\cos x + 1 \:=\:0\quad\Rightarrow\quad \cos x \:=\:-\frac{1}{2}\quad\Rightarrow\quad\boxed{ x \;=\;\frac{2\pi}{3},\:\frac{4\pi}{3}}$