How do I solve this? \cos 3x=\cos^2 x-3 \sin^2 x

Alan Smith

Alan Smith

Answered question

2021-12-30

How do I solve this?
cos3x=cos2x3sin2x

Answer & Explanation

Karen Robbins

Karen Robbins

Beginner2021-12-31Added 49 answers

If you write cos(3x)=cos(x+2x), you can then apply the identity for the cosine of a sum. After that you can apply the identities that reduce cos(2x) and sin(2x) to functions of cosx and sinx. You'll end up with an expression in which all sines are squared, so you can apply the identity sin2x=1cos2x. You're left with no trigonometric expressions except cosx. Then the substitution u=cosx reduces it to an algebraic equation to be solved for u.
You shouldn't wait until you know how to solve it before you start working on it. If you start with cos(x+2x) and apply the identity for the cosine of a sum, you can just see where that takes you. That's how I did this.
aquariump9

aquariump9

Beginner2022-01-01Added 40 answers

cos(3)x=4cos3x3cosx,  4cos2x3=cos2x3sin2x
hence by setting z=cosx we have to solve:
4z34z23z+3=(z1)(4z23)=0
so cosx=1 or cosx=32, from which x=2kπ or x=±π6+2kπ
nick1337

nick1337

Expert2022-01-08Added 777 answers

Using

cos3x=cos3x3cosxsin2x,cos3x=cos2x3sin2xcos3x3cosxsin2x=cos2x3sin2x(cosx)(cos2x3sin2x)=cos2x3sin2xcosx=1   or  tan2x=13

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