Solving Trig Equations (Half Angle/Double Angle Formulas)

[Frubens]

Hey, I have a big test tomorrow and I've really been struggling. If anyone could provide any assistance I'd really appreciate it. I have a whole plethora of problems I could list but here are just a few and hopefully I'll get the gist of it.

SOLVE on x is greater than or equal to 0 and less than 2 pi.

3tan^2(x) - 1 = 0

Here's what I've tried:
3tan^2(x) = 1
tan^2(x) = 1/3
I really don't know what to do at this point...I've tried things like...
(1 - cos(2x))/(1+cos(2x)) = 1/3
But I just don't know where to go with it at this point.

Another problem.

sin^2(x) - 2sin(x)cos(x) = cos^2(x)

Here's what I've tried:
sin^2(x) - 2sin(x)cos(x) - cos^2(x) = 0

I feel like I should be able to factor that and then continue but I don't know >_>.

Any help is appreciated. Thanks.

 

[srmichael]

Hey, I have a big test tomorrow and I've really been struggling. If anyone could provide any assistance I'd really appreciate it. I have a whole plethora of problems I could list but here are just a few and hopefully I'll get the gist of it.

SOLVE on x is greater than or equal to 0 and less than 2 pi.

3tan^2(x) - 1 = 0

Here's what I've tried:
3tan^2(x) = 1
tan^2(x) = 1/3
I really don't know what to do at this point...I've tried things like...
(1 - cos(2x))/(1+cos(2x)) = 1/3
But I just don't know where to go with it at this point.

Another problem.

sin^2(x) - 2sin(x)cos(x) = cos^2(x)

Here's what I've tried:
sin^2(x) - 2sin(x)cos(x) - cos^2(x) = 0

I feel like I should be able to factor that and then continue but I don't know >_>.

Any help is appreciated. Thanks.

For the first once you get \(\displaystyle \tan^2(x) = \frac{1}{3}\) then you can take the square root of both sides. Thus you would get

\(\displaystyle \tan(x) = \pm\frac{1}{\sqrt{3}}\)

Now you can find the angles that satisfy this.

For the second one, use these identities: \(\displaystyle \cos^2(x) - \sin^2(x) = \cos(2x)\) and \(\displaystyle 2\sin(x)\cos(x) = \sin(2x)\). Therefore,

\(\displaystyle \sin^2(x) - 2\sin(x)\cos(x) - \cos^2(x) = 0\) becomes

\(\displaystyle -[\cos^2(x) + 2\sin(x)\cos(x) - \sin^2(x)] = 0\) which becomes

\(\displaystyle -[\cos(2x) + \sin(2x)] = 0\) Thus,

\(\displaystyle \cos(2x)+\sin(2x) = 0\)

\(\displaystyle \cos(2x) = -\sin(2x)\)

\(\displaystyle \frac{\cos(2x)}{\cos(2x)} = -\frac{\sin(2x)}{\cos(2x)}\)

\(\displaystyle 1 = -\tan(2x)\)

From here you can solve for x