Help solve a trigonometric equation,

Question

$$3\sin^2(x)+5\sin( x)\cos( x) = 2$$

Can't seem to solve it. Anyone can help me?

Answer 1

Hint: You can use the double angle formulas to make it a linear equation in $\sin 2x$ and $\cos 2x$. Then you can use the sum formula to make it $A\cos (2x+\phi).$

Answer 2

Divide both sides by $\cos^2x$ to get $3\tan^2x+5\tan x=2\sec^2x$ and then use the identity $\sec^2x=1+\tan^2x$ to make the equation

$$3\tan^2x+5\tan x=2+2\tan^2x$$

or

$$t^2+5t-2=0$$

where $t=\tan x$. The rest of the (rather ugly) solution is as given by Américo Tavares, except I would simply write things as

$$x=\arctan\left({-5\pm\sqrt{33}\over2}\right)+k\pi\quad\text{for } k\in\mathbb{Z}$$

(Note, this solution is essentially the same as Fly by Night's, I just got to the quadratic by a slightly different route. I should have noted that it's OK to divide by $\cos^2x$ since the original equation cannot hold if $\cos x=0$.)

Answer 3

I'd prefer to represent $2$ as $2(\cos^2x+\sin^2x)$. And then divde the left and the right parts of the equation by $\cos^2x$. Then simple replacement leads to quadratic equation...

Answer 4

Since $\cos^2x + \sin^2x \equiv 1$ we have $2\cos^2x + 2\sin^2x \equiv 2$. Hence: \begin{eqnarray*} 3\sin^2x + 5\sin x \cos x = 2 &\iff& 3\sin^2x + 5\sin x \cos x = 2\sin^2+2\cos^2 x \\ \\ &\iff& \sin^2x + 5\sin x\cos x - 2\cos^2x = 0 \\ \\ &\iff& (\tan^2x + 5\tan x - 2)\cos^2x = 0 \end{eqnarray*} Notice that when $\cos x =0$ we have $\sin x = \pm 1$ and hence $3\sin^2x + 5\sin x \cos x = 3 \neq 2$. It follows that no solution of $3\sin^2x + 5\sin x \cos x = 2$ has $\cos x =0$. The only solutions have $$\tan^2x + 5\tan x - 2=0$$ This is a quadratic in $\tan x$. We have $u^2 + 5u - 2 = 0 \iff u = -\frac{5}{2} \pm \frac{1}{2}\sqrt{33}$. Hence: $$\tan^2x + 5\tan x - 2=0 \iff x = \arctan\left(-\frac{5}{2} \pm \frac{1}{2}\sqrt{33}\right).$$