# Solving Quandratic Trigonometric Equations

#### misiaizeska

How would I solve these equations?
1) (2sinx - 1)cosx = 1
2) (2 cosx + √3)sinx = 0

I tried to convert them to the same identity, but wasn't able to find a substitute in neither the pythagorean identities or the double angle formulas, and I'm lost as to what to do now. Any suggestions/advice would be appreciated, thanks in advance!

#### Soroban

MHF Hall of Honor
Hello, misiaizeska!

$$\displaystyle [1]\;(2\sin x - 1)\cos x \:=\: 1$$

$$\displaystyle [2]\;(2\cos x + \sqrt{3})\sin x \:=\: 0$$

The second one is easy . . . the product is zero.

We have:
. . $$\displaystyle 2\cos x+\sqrt{3} \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-\tfrac{\sqrt{3}}{2} \quad\Rightarrow\quad x \:=\:\begin{Bmatrix}\frac{5\pi}{6} + 2\pi n \\ \frac{7\pi}{6} + 2\pi n \end{Bmatrix}$$
. . $$\displaystyle \sin x \:=\:0 \quad\Rightarrow\quad x \:=\:\pi n$$

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I have an incomplete solution for the first one.

We have: .$$\displaystyle 2\sin x\cos x - \cos x \:=\:1 \quad\Rightarrow\quad 2\sin x\cos x \:=\:1 + \cos x$$

Square both sides: . . . . . . . $$\displaystyle 4\sin^2\!x\cos^2\!x \:=\:1 + 2\cos x + \cos^2\!x$$

. . . . . . . . . . . . . . . . . $$\displaystyle 4(1-\cos^2\!x)\cos^2x \:=\:1 + 2\cos x + \cos^2\!x$$

. . . . . . . . . . . . . . . . . . $$\displaystyle 4\cos^2x - 4\cos^4\!x \:=\:1 + 2\cos x + \cos^2\!x$$

n . . . . . . . . $$\displaystyle 4\cos^4\!x- 3\cos^2\!x + 2\cos x + 1 \:=\:0$$

. . $$\displaystyle (\cos x + 1)(4\cos^3\!x - 4\cos^2x + \cos x + 1) \:=\:0$$

I found one root: .$$\displaystyle \cos x + 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-1 \quad\Rightarrow\quad x \:=\:\pi + 2\pi n$$

I don't know if there are any others . . .

2 people

#### misiaizeska

Thank you very much, that was really helpful and I was able to figure it out!