Solve the trigonometric equation.
√2cos(x5−π12)−√6sin(x5−π12)=2(sin(x5−2π3)−sin(3x5+π6))
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 \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*}
\displaystyle 4 \cos \left(2A-\frac{\pi}{12}\right)\cos A=\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 of:
1.
\displaystyle \cos 2A=\frac{1-\tan^2 A}{1+\tan^2 A},
2.
\displaystyle \sin 2A=\frac{2\tan A}{1+\tan^2 A}, 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:
\displaystyle \dfrac{4\cos \left(2A-\frac{\pi}{12}\right)\cos A}{\cos A}=\dfrac{\sqrt{6}\sin A- \sqrt{2}\cos A}{\cos A}
\displaystyle 4\cos \left(2A-\frac{\pi}{12}\right)=\sqrt{6}\tan A- \sqrt{2}
\displaystyle 4\cos 2A \cos \left(\frac{\pi}{12}\right)+4\sin 2A \sin \left(\frac{\pi}{12}\right)=\sqrt{6}\tan A- \sqrt{2}
\displaystyle \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}
\displaystyle \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)
\displaystyle \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)
\displaystyle \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)
\displaystyle \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)
\displaystyle \sqrt{3}\tan^3 A+\sqrt{3}\tan^2 A+(2-\sqrt{3})\tan A-(2+\sqrt{3})=0 (*)
It's quite obvious that \displaystyle \tan A=1 is one of the solution to (*) since:
\displaystyle \begin{align*}\sqrt{3}+\sqrt{3}+(2-\sqrt{3})-(2+\sqrt{3})&=2\sqrt{3}+2-\sqrt{3}-2-\sqrt{3}\\&=0\end{align*}
We then use the long division to find the other two roots.
\displaystyle (\tan A-1)(\sqrt{3}\tan^2 A+2\sqrt{3}\tan A+2+\sqrt{3})=0
Since the discriminant of the quadratic expression that we found above \sqrt{3}\tan^2 A+2\sqrt{3}\tan A+2+\sqrt{3}:
\displaystyle \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*}
is a negative value:, we can say the other two roots are imaginary.
Hence the solutions that we have found are
\displaystyle \tan A=1 which gives the A as
\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.
I just loved the beauty of this question...
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