Thread: Solve the double angle equation

Im stuck on this question in my maths homework:

Solve the equation √3tan2x-1=0 for 0<X<π

Here is what i've done:

√3tan2x-1=0
tan2x=1/√3
Using trig identities,
tanx= sinx/cosx
so tan2x=sin2x/cos2x < is this right?
I've then used the trig expansions:
2sinxcosx/1-2sin^2x < Am I using the right expansion for cos2x?

Am I using the right method? I have tried to go further than this, but nothing has worked out, so where do I go from here?

Thanks in advance.

Don't bother changing it to sin and cos - arctan is a fine operation and you can use it on your second line



That is the principle solution but remember that since you have a double angle the period is so you'll need to add that to get the second solution

I would do that, but we havent been taught it in class and its not in our notes, so I have to do it the sin and cos way. Thanks anyway.

Originally Posted by Fionafaye

I would do that, but we havent been taught it in class and its not in our notes, so I have to do it the sin and cos way.

Draw an equilateral triangle with sides of length two.
Draw an altitude from the summit angle.
It is also a median and angular bisector.
The altitude has length and the adjunct side has length .
So half the summit angle is and .
Now finish.

If you want it in sin and cos, then

√3sin(2x) = cos(2x)

sin^2(x) + 2√3sin(x)cos(x) - cos^2(x) = 0

Solve the quadratic for sin(x). From that find tan(x)

a tricky kind of question.