Solving Quandratic Trigonometric Equations

**misiaizeska** · January 9th 2013, 08:09 PM

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** · January 9th 2013, 09:35 PM

Hello, misiaizeska!

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

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

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

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

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I have an incomplete solution for the first one.

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

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

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

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

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

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

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

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

**misiaizeska** · January 11th 2013, 02:48 AM

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

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