Thread: 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.

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

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

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,



When n is even,



When n is odd,



Hence,

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.