# Math Help - Solving trig equations

1. ## Solving trig equations

Ok Well, I need a lot of help with this. I hav eso many problems to do and I do not even know where to start.

I just want to get help with a few and then I'll see if I can figure out how to do them on my own..

Ok so here is the first type of problem I have.

Solve each trig equation.

First off I'm not entirely sure what I'm trying to find in the following. Even then I don't really know where to start.
1. sin x = -sqrt2/2

2. Cos 2x=-sqrt2/2

3. tan .5x= sqrt3

4. sqrt3 sec x +2 =0

I think if i know how to do those I will be able to figure out the rest of those type of problems.

Here id the other type of problem i have to do.

Solve the each trig equation

1.4 sin^2 x-3=0

For this one would I just add 3 then divide by 4. Then take the inverse sin of sqrt of 3/4 and the answer would be 1.047. If that is right then that would be one of the solutions but Im not sure how to find the other one

2. cos x +2= 3cos x

3. 2 sin ^2 x-3=0

2. Originally Posted by mtt0216
Ok Well, I need a lot of help with this. I hav eso many problems to do and I do not even know where to start.

I just want to get help with a few and then I'll see if I can figure out how to do them on my own..

Ok so here is the first type of problem I have.

Solve each trig equation.

First off I'm not entirely sure what I'm trying to find in the following. Even then I don't really know where to start.
1. sin x = -sqrt2/2

2. Cos 2x=-sqrt2/2

3. tan .5x= sqrt3

4. sqrt3 sec x +2 =0

I think if i know how to do those I will be able to figure out the rest of those type of problems.

Here id the other type of problem i have to do.

Solve the each trig equation

1.4 sin^2 x-3=0

For this one would I just add 3 then divide by 4. Then take the inverse sin of sqrt of 3/4 and the answer would be 1.047. If that is right then that would be one of the solutions but Im not sure how to find the other one

2. cos x +2= 3cos x

3. 2 sin ^2 x-3=0
A) For the first lot you can take the inverse function.

For $\sin(x) = -\frac{\sqrt2}{2}$

$x = \arcsin \left(-\frac{\sqrt2}{2}\right)$

You should also know your special triangles

B) Divide by 4. Then use the difference of two squares

$\left(\sin (x) - \frac{\sqrt3}{2}\right) \left(\sin (x) + \frac{\sqrt3}{2}\right)=0$

Which will give two solutions

3. Well I'm still confused. in my class we have never used arcsin so I don't even know what that is or how to use it.

And for the second part I'm not exactly sure what you did.

4. ## 2

arcsin basically means $sin^{-1}\theta$ which is an inverse function

#2

$\cos{(2x)}=-\frac{\sqrt{2}}{2}$
$
\cos^{-1}\left({-\frac{\sqrt{2}}{2}}\right) = 2x$

$\frac{3\pi}{4} = 2x$

$\frac{3\pi}{8} = x$

5. ..........

6. thanks for the help guys.

I did some of the problems and I wanted to see if i got the right answer for them.

so

$sin=-\sqrt{2} /2$

x= $\frac{5pi}{4}$ or $\frac{7pi}{4}$

==========================
cos 2x= $\frac{\sqrt{2}}{2}$

x= $\frac{3pi}{8}$ or $\frac{5pi}{8}$

===========================

$sin(x+\frac{pi}{4} )$= $\frac{-\sqrt{2} }{2}$

x= $6\frac{pi}{4}$ or $4\frac{pi}{4}$

=========================
$
tan.5x=\sqrt{3}$

$x= \frac{pi}{1.5}$
=========================
$2sinx+\sqrt{3}=0$

$x= 5\frac{pi}{3}$ or $4\frac{pi}{3}$

=============================

$4 tanx-2=2$

$x= \frac{pi}{4}$

I wasn't able to get these 2.
$tan(x-30deg)=\sqrt{3}$

$\sqrt{3} secx+2=0$

Also how do I show that an equation has an infinite amount of solutions