Solving Quandratic Trigonometric Equations

Sep 2012
28
0
canada
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

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May 2006
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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 . . .
 
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Sep 2012
28
0
canada
Thank you very much, that was really helpful and I was able to figure it out!