# [SOLVED] A challenging trigonometric equation?

• June 23rd 2008, 06:26 AM
fardeen_gen
[SOLVED] A challenging trigonometric equation?
Solve √(√3 cos x + sin x - 2) + √(cot 3x + sin^2 x - 1/4) = sin (3x/2) + √2/2?

^- raised to
Answer: (24n + 13)π/6 , n belongs to Z

How to solve this? My answer is not matching.

Any help would be appreciated.
• June 23rd 2008, 12:00 PM
red_dog
The equation can be written as
$\displaystyle\sqrt{2}\sqrt{\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x-1}+\sqrt{\cot 3x+\sin^2x-\frac{1}{4}}=\sin\frac{3x}{2}+\frac{\sqrt{2}}{2}\L eftrightarrow$
$\displaystyle\Leftrightarrow\sqrt{2}\sqrt{\sin\lef t(x+\frac{\pi}{3}\right)-1}+\sqrt{\cot 3x+\sin^2x-\frac{1}{4}}=\sin\frac{3x}{2}+\frac{\sqrt{2}}{2}$
We have to put the condition $\sin\left(x+\frac{\pi}{3}\right)\geq 1$. But $\sin\left(x+\frac{\pi}{3}\right)\leq 1, \ \forall x\in\mathbf{R}$.
Then $\displaystyle\sin\left(x+\frac{\pi}{3}\right)=1\Ri ghtarrow x=\frac{\pi}{6}+2k\pi, \ k\in\mathbf{Z}$.
Now $\displaystyle \cot3\left(\frac{\pi}{6}+2k\pi\right)=0, \ \sin^2\left(\frac{\pi}{6}+2k\pi\right)=\frac{1}{4}$
so, the left hand member of the equation is equal to 0.
Then $\displaystyle \sin\frac{3}{2}\left(\frac{\pi}{6}+2k\pi\right)=-\frac{\sqrt{2}}{2}\Rightarrow x=\frac{(24p+13)\pi}{6}, \ p\in\mathbf{Z}$
• June 27th 2008, 08:59 PM
fardeen_gen
Can anyone explain the last step?
• June 28th 2008, 12:42 AM
red_dog
$\displaystyle\sin\left(\frac{\pi}{4}+3k\pi\right)=-\frac{\sqrt{2}}{2}\Rightarrow\sin\left(\frac{\pi}{ 4}+k\pi\right)=-\frac{\sqrt{2}}{2}$
Then $k$ must be odd: $k=2p+1$
Now $\displaystyle x=\frac{\pi}{6}+2k\pi=\frac{\pi}{6}+2(2p+1)\pi=\fr ac{(24p+13)\pi}{6}$