Efficient Trigonometric Identity Solution for Finding tan x

[teleport]

Homework Statement

Find tan x if

\dfrac {\sin^2 x}{3} + \dfrac {\cos^2 x}{7} = \dfrac {-\sin(2x) + 1}{10}

Homework Equations

Trigonometric identities.

The Attempt at a Solution

I have tried removing the cos squared on the LHS by using

\cos^2 x + \sin^2 x = 1

and then using

\sin^2 x = \dfrac {1 - \cos(2x)}{2}

then using

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

Noticing the RHS equals

(\sin x - \cos x)^2

and factoring with what is now the LHS, would eventually give me an expression similar to

a\sin(2x) + b\cos(2x) = c

where a,b,c are constants. If there is now a way to solve for any of sin(2x) or cos(2x), I can then get tan(2x) and hence, tanx.

This solution has been so long, that it all sounds very suspicious to me. I have a hunch there should be a much easier solution. Thanks for any help, but please, don't post a solution.

 

[VietDao29]

Homework Statement

Find tan x if

\dfrac {\sin^2 x}{3} + \dfrac {\cos^2 x}{7} = \dfrac {-\sin(2x) + 1}{10}

Your method looks okay. But if you are searching for a better one, you may consider dividing both sides by cos²(x), since it's trivial to see that cos(x) = 0 is not a solution to the equation.

It becomes:
\frac{\tan ^ 2 x}{3} + \frac{1}{7} = -2 \frac{\tan x}{10} + \frac{1}{10 \cos ^ 2 (x)}

\Leftrightarrow \frac{\tan ^ 2 x}{3} + \frac{1}{7} = - \frac{\tan x}{5} + \frac{1}{10} \left( 1 + \tan ^ 2 x \right)

Can you go from here? :) Hint: It's a quadratic.

 

[teleport]

Ouch! That hurts. I didn't even think of dividing by (cosx)^2, otherwise I would have seen it right away. It was right in my face all the time. If it were a shark, I would be ...