# Thread: general solution of trig equation

1. ## general solution of trig equation

Every things is alright now, but the reference answer is n*pi+[(-1)^n-1]*pi/4, which is different from mine, could anyone tell me how to combine that 2 solutions I found together?

Again, a little problem, I found that the second general solution is strange because it make sec 5x + csc 2x undefined in some cases of n.

2. ## Re: general solution of trig equation

Honestly, I don't understand your question. Is this what you have done? Can you be more specific about what's your solutions etc ... ?

3. ## Re: general solution of trig equation

sorry for that...I have posted the wrong picture, now corrected

for the first question, I just want to know how to find the solution which is n*pi+[(-1)^n-1]*pi/4, where n is an integer. Is there something wrong with my process?

for the second question, I think I have solved the equation correctly, but the solution (2n*pi/3 - pi/2) can't satisfy the equation when n=3..etc

4. ## Re: general solution of trig equation

Originally Posted by piscoau

Every things is alright now, but the reference answer is n*pi+[(-1)^n-1]*pi/4, which is different from mine, could anyone tell me how to combine that 2 solutions I found together?
Hi piscoau,

For the first problem your answer is correct and the given answer is an equivalent. For,

$x=n\pi+\left[(-1)^n-1\right]\frac{\pi}{4}\mbox{ where }n\in Z$

When n is even, $n=2k~;~k\in Z$

$x=2k\pi+\left[(-1)^{2k}-1\right]\frac{\pi}{4}=2k\pi\mbox{ where }k\in Z$

When n is odd, $n=2k+1~;~k\in Z$

$x=(2k+1)\pi+\left[(-1)^{2k+1}-1\right]\frac{\pi}{4}=(2k+1)\pi-\frac{\pi}{2}=2k\pi+\frac{\pi}{2}\mbox{ where }k\in Z$

Hence, $x=2k\pi\mbox{ or }x=2k\pi+\frac{\pi}{2}\mbox{ where }k\in Z$

Originally Posted by piscoau
for the second question, I think I have solved the equation correctly, but the solution (2n*pi/3 - pi/2) can't satisfy the equation when n=3..etc
Yes you have solved it correctly and it satisfies the equation for n=3.