Find the general solution of \(\) My approach:

Axel Stout

Axel Stout

Answered question

2022-04-13

Find the general solution of
My approach: Converted them to cosines and then further simplified. I got or But my book stated the answer as or Where am i going wrong.
My simplification:
cos2θcos4θ=2cos4θcos2θ
cos2θcos4θ=cos6θ+cos2θ
cos4θ=cos6θ
cos4θ=cos(π6θ)
4θ=2nπ±(π6θ)
taking the plus and minus, I got my above-stated answer. But it doesn't quite agree with the answer given in my book. Can anyone explain my mistake?

Answer & Explanation

riasc31lj

riasc31lj

Beginner2022-04-14Added 8 answers

You got the same answer, just in a different form
±π2+2nπ=π2,π2,3π2,5π2, =π2+nπ
±π10+2nπ5=π10,π10,3π10,5π10, =π10+nπ5
glanzerjbdo

glanzerjbdo

Beginner2022-04-15Added 13 answers

Hint:
Just another approach:
Prosthaphaeresis Formula:
cos4θ+cos6θ=2cosθcos5θ
Now if cos5θ=0,5θ=(2n+1)π2 where n is any integer
Observe that the solution of cosθ=0 is a subset of the previous one.

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