MHB Solve A Trigonometric Equation

AI Thread Summary
The discussion centers on solving the trigonometric equation involving cosine and sine functions with specific arguments. A user corrected a typo in the original problem, clarifying that the second term should be a sine function rather than a cosine function. There is a consensus that the arguments on the left side of the equation should match. The conversation highlights the importance of accuracy in mathematical expressions to avoid confusion. Overall, the thread emphasizes collaborative problem-solving in trigonometry.
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$$\sqrt{2}\cos \left(\frac{x}{5}-\frac{\pi}{12} \right)-\sqrt{6}\sin \left(\frac{x}{5}-\frac{\pi}{12} \right)=2\left(\sin \left(\frac{x}{5}-\frac{2\pi}{3} \right)-\sin \left(\frac{3x}{5}+\frac{\pi}{6} \right) \right)$$
 
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Did you want the arguments on both cosine terms to read the same?
 
Jester said:
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)
 
My solution:

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

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

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

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

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

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

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

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

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

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

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

Hence the solutions are

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

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