Tuesday, June 16, 2015

Hardest Trigonometric Equation (First Solution)

Solve the trigonometric equation.

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

My solution:

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

[MATH]\begin{align*}\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)\\&=2\left(2\cos \left(2A-\frac{\pi}{12} \right)\sin \left(-A-\frac{\pi}{2} \right) \right)\\&=-4\left(\cos \left(2A-\frac{\pi}{12} \right)\sin \left(A+\frac{\pi}{2} \right) \right)\\&=-4  \cos \left(2A-\frac{\pi}{12}\right)\cos A \end{align*}[/MATH]


[MATH]4  \cos \left(2A-\frac{\pi}{12}\right)\cos A=\sqrt{6}\sin A- \sqrt{2}\cos A[/MATH]

Now, divide the left and right side of the equation by [MATH]\cos A[/MATH] and use the formulas of:

1.

[MATH]\cos 2A=\frac{1-\tan^2 A}{1+\tan^2 A}[/MATH],

2.

[MATH]\sin 2A=\frac{2\tan A}{1+\tan^2 A}[/MATH], and

3.

$\sin \left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}}{4}(\sqrt{3}-1)$;

$\cos \left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}}{4}(\sqrt{3}+1)$


to further simplify the equation, we get:

[MATH]\dfrac{4\cos \left(2A-\frac{\pi}{12}\right)\cos A}{\cos A}=\dfrac{\sqrt{6}\sin A- \sqrt{2}\cos A}{\cos A}[/MATH]

[MATH]4\cos \left(2A-\frac{\pi}{12}\right)=\sqrt{6}\tan A- \sqrt{2}[/MATH]

[MATH]4\cos 2A \cos \left(\frac{\pi}{12}\right)+4\sin 2A \sin \left(\frac{\pi}{12}\right)=\sqrt{6}\tan A- \sqrt{2}[/MATH]

[MATH]\small 4\left(\frac{1-\tan^2 A}{1+\tan^2 A}\right) \left(\dfrac{\sqrt{2}}{4}(\sqrt{3}+1)\right)+4\left(\frac{2\tan A}{1+\tan^2 A}\right) \left(\dfrac{\sqrt{2}}{4}(\sqrt{3}-1)\right)=\sqrt{6}\tan A- \sqrt{2}[/MATH]

[MATH]\scriptsize \cancel{4}\left(\frac{1-\tan^2 A}{1}\right) \left(\dfrac{\sqrt{2}}{\cancel{4}}(\sqrt{3}+1)\right)+\cancel{4}\left(\frac{2\tan A}{1}\right) \left(\dfrac{\sqrt{2}}{\cancel{4}}(\sqrt{3}-1)\right)=(\sqrt{6}\tan A- \sqrt{2})(1+\tan^2 A)[/MATH]

[MATH]\small \left(1-\tan^2 A\right) \left(\sqrt{2}(\sqrt{3}+1)\right)+\left(2\tan A\right) \left(\sqrt{2}(\sqrt{3}-1)\right)=\sqrt{2}(\sqrt{3}\tan A- 1)(1+\tan^2 A)[/MATH]

[MATH]\small \left(1-\tan^2 A\right) \left(\cancel{\sqrt{2}}(\sqrt{3}+1)\right)+\left(2\tan A\right) \left(\cancel{\sqrt{2}}(\sqrt{3}-1)\right)=\cancel{\sqrt{2}}(\sqrt{3}\tan A- 1)(1+\tan^2 A)[/MATH]

[MATH]\left(1-\tan^2 A\right) (\sqrt{3}+1)+\left(2\tan A\right)(\sqrt{3}-1)=(\sqrt{3}\tan A- 1)(1+\tan^2 A)[/MATH]

[MATH]\sqrt{3}\tan^3 A+\sqrt{3}\tan^2 A+(2-\sqrt{3})\tan A-(2+\sqrt{3})=0[/MATH] (*)

It's quite obvious that [MATH]\tan A=1[/MATH] is one of the solution to (*) since:

[MATH]\begin{align*}\sqrt{3}+\sqrt{3}+(2-\sqrt{3})-(2+\sqrt{3})&=2\sqrt{3}+2-\sqrt{3}-2-\sqrt{3}\\&=0\end{align*}[/MATH]

We then use the long division to find the other two roots.

[MATH](\tan A-1)(\sqrt{3}\tan^2 A+2\sqrt{3}\tan A+2+\sqrt{3})=0[/MATH]

Since the discriminant of the quadratic expression that we found above $\sqrt{3}\tan^2 A+2\sqrt{3}\tan A+2+\sqrt{3}$:

[MATH]\begin{align*}\text{discriminant}&=(-2\sqrt{3})^2-4(\sqrt{3})(2+\sqrt{3})\\&=12-12-8\sqrt{3}\\&=-8\sqrt{3}\\&\lt 0\end{align*}[/MATH]

is a negative value:, we can say the other two roots are imaginary.

Hence the solutions that we have found are

[MATH]\tan A=1[/MATH] which gives the $A$ as

[MATH]A=\frac{\pi}{4}+2n\pi[/MATH] where n is an integer, i.e.

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


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