How do I solve this? cos⁡3x=cos2x−3sin2x

Question

How do I solve this?  cos⁡3x=cos2x−3sin2x

Karen Robbins · Accepted Answer

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&#039;ll end up with an expression in which all sines are squared, so you can apply the identity sin2x=1−cos2x. You&#039;re left with no trigonometric expressions except cos⁡x. Then the substitution u=cos⁡x reduces it to an algebraic equation to be solved for u.You shouldn&#039;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&#039;s how I did this.

aquariump9 · Accepted Answer

cos⁡(3)x=4cos3x−3cos⁡x,  4cos2x−3=cos2x−3sin2x  hence by setting z=cos⁡x we have to solve:  4z3−4z2−3z+3=(z−1)(4z2−3)=0  so cos⁡x=1 or cos⁡x=32, from which x=2kπ or x=±π6+2kπ

nick1337 · Accepted Answer

Using  cos⁡3x=cos3x−3cos⁡xsin2⁡x,cos⁡3x=cos2⁡x−3sin2⁡xcos3⁡x−3cos⁡xsin2⁡x=cos2⁡x−3sin2⁡x(cos⁡x)(cos2⁡x−3sin2⁡x)=cos2⁡x−3sin2⁡xcos⁡x=1   or  tan2⁡x=13

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

Answered question

Answer & Explanation

New Questions in Trigonometry