Solve $6\cos^2(x) - 8\cos(x)\sin(x) + 1 = 0$

Question

How would you solve this equation for answers between $0$ and $360$ degrees? $6\cos^2(x) - 8\cos(x)\sin(x) + 1 = 0$

(edit)I have tried the following:

• $10\cos(x)(6/10\cos(x) - 8/10\sin(x)) + 1 = 0 $
• $10\cos(x)(\cos(y)\cos(x) - \sin(y)(\sin(x)) + 1 = 0$
• $10\cos(x)\cos(x+y) + 1 = 0$

Answer 1

Hint:

Recall that $2\cos^2 x=1+\cos (2x)$ and $2\cos x\sin x=\sin (2x).$

Answer 2

hint: Use the so-called Weierstrass substition: $$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$

Answer 3

Hint:

As $\sin x\cos x\ne0$

Divide both sides by $\cos^2x$ to form a quadratic equation in $\tan x$

Or divide both sides by $\sin^2x$

Or use double angle formula for $\sin2x,\cos2x$

Answer 4

I like the following way.

We have $$6\cos^2x-8\sin{x}\cos{x}+\sin^2x+\cos^2x=0$$ or $$\sin^2x-8\sin{x}\cos{x}+7\cos^2x=0$$ or $$(\sin{x}-\cos{x})(\sin{x}-7\cos{x})=0.$$ 1. $\sin{x}=\cos{x}.$

$$x=45^{\circ}+180^{\circ}k$$ where $k\in\{0,1\}$.

2. $\sin{x}=7\cos{x},$ which gives $$x=\arctan7+180^{\circ}k,$$ where $k\in\{0,1\}$.