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Solve A Trigonometric Equation

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anemone

MHB POTW Director
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Feb 14, 2012
3,685
Solve

\(\displaystyle \sqrt{2}\cos \left(\frac{x}{5}-\frac{\pi}{12} \right)-\sqrt{6}\sin \left(\frac{x}{5}-\frac{\pi}{12} \right)=2\left(\sin \left(\frac{x}{5}-\frac{2\pi}{3} \right)-\sin \left(\frac{3x}{5}+\frac{\pi}{6} \right) \right)\)
 

Jester

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MHB Math Helper
Jan 26, 2012
183
Did you want the arguments on both cosine terms to read the same?
 

MarkFL

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Feb 24, 2012
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Did you want the arguments on both cosine terms to read the same?
I have edited anemone's post to correct the typo. She is offline now, but I am aware of the problem from another site, so I have taken the liberty to change the statement of the problem, and I will be refraining from posting a solution as I have seen the problem before.

The arguments on the left are supposed to be the same, however the second term was meant to be a sine function instead.

On behalf of anemone, I want to thank you for catching this! (Nod)
 
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anemone

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Feb 14, 2012
3,685
My solution:

By letting \(\displaystyle A=\frac{x}{5}-\frac{\pi}{12}\) use the sum-to-product formula to simplify the LHS of the equation, I get:

\(\displaystyle \sqrt{2}\cos A-\sqrt{6}\sin A=2\left(\sin \left(A-\frac{7\pi}{12} \right)-\sin \left(3A+\frac{5\pi}{12} \right) \right)\)

\(\displaystyle \sqrt{2}\cos A-\sqrt{6}\sin A=2\left(2\cos \left(2A-\frac{\pi}{12} \right)\sin \left(-A-\frac{\pi}{2} \right) \right)\)

\(\displaystyle \sqrt{2}\cos A-\sqrt{6}\sin A=-4\left(\cos \left(2A-\frac{\pi}{12} \right)\sin \left(A+\frac{\pi}{2} \right) \right)\)

\(\displaystyle \sqrt{2}\cos A-\sqrt{6}\sin A=-4 \left( \cos \left(2A-\frac{\pi}{12}\right)\right)\left( \cos A \right)\)

\(\displaystyle 4\cos A \left( \cos \left(2A-\frac{\pi}{12}\right)\right)=\sqrt{6}\sin A- \sqrt{2}\cos A\)

Now, divide the left and right side of the equation by \(\displaystyle \cos A\) and use the formulas for \(\displaystyle \cos 2A=\frac{1-\tan^2 A}{1-\tan^2 A}\) and \(\displaystyle \sin 2A=\frac{2\tan A}{1-\tan^2 A}\) to further simplify the equation yields:

\(\displaystyle \sqrt{6}\tan^3 A+\sqrt{6}\tan^2 A+(2\sqrt{2}-\sqrt{6})\tan A-(2\sqrt{2}+\sqrt{6})=0\) (*)

It's quite obvious that \(\displaystyle \tan A=1\) is one of the solution to (*) and use the long division to find the other two roots.

\(\displaystyle (\tan A-1)(\sqrt{6}\tan^2 A+2\sqrt{6}\tan A+2\sqrt{2}+\sqrt{6})=0\)

Since the discriminant of the quadratic expression that we found above is a negative value (\(\displaystyle -8\sqrt{12}\)), we can say the other two roots are imaginary roots.

Hence the solutions are

\(\displaystyle A=\frac{\pi}{4}+2n\pi\) where n is an integer, i.e.

\(\displaystyle \frac{x}{5}-\frac{\pi}{12}=\frac{\pi}{4}+2n\pi\) which gives us \(\displaystyle x=\frac{5\pi}{3}+10n\pi\).