Solve $5\sin^2(x) + \sin(2x) - \cos^2(x) = 1$

Question

I tried $$5\sin^2(x) + 2\sin(x)\cos(x)- (1-\sin^2(x)) = 1.$$

Simplifying,

$$6\sin^2(x) + 2\sin(x)\cos(x) -2 = 0$$

Then I'm stuck!

Answer 1

Using that $$-2 = -2\sin^2(x) - 2\cos^2(x)$$ from $$6\sin^2(x) + 2\cos(x)\sin(x) - 2 = 0$$ we get $$4\sin^2(x) + 2\cos(x)\sin(x) - 2\cos^2(x) = 0$$ and dividing by $\cos^2(x)$ we have $$4\tan^2(x) + 2\tan(x) - 2 = 0$$

Can you continue from here? Note that when dividing by $\cos^2(x)$ we assume that it is not $0$. What if it is $0$?

Answer 2

Hint

Use the double angle formula $\cos(2x)=1-2 \sin(x)^2$; extract from it $\sin(x)^2$ and replace. You will end with $$\sin (2 x)-3 \cos (2 x)+1=0$$

I am sure that you can take from here.

Answer 3

Using the twice-angle formulas, this becomes $$ -3 \cos(2x) + \sin(2x) = -1 $$ Let $\alpha = \arctan(1/3)$ so $\cos(\alpha) = 3/\sqrt{10}$ and $\sin(\alpha) = 1/\sqrt{10}$. The equation can be written as $$ - \cos(\alpha) \cos(2x) + \sin(\alpha) \sin(2x) = - 1/\sqrt{10}$$ or $$\cos(\alpha + 2 x) = 1/\sqrt{10}$$

Answer 4

$\cos x = \frac{2}{\sqrt5}$ therefore $x=26.56$ in the first quadrant