Alan Smith

2021-12-30

How do I solve this?
$\mathrm{cos}3x={\mathrm{cos}}^{2}x-3{\mathrm{sin}}^{2}x$

Karen Robbins

If you write $\mathrm{cos}\left(3x\right)=\mathrm{cos}\left(x+2x\right)$, you can then apply the identity for the cosine of a sum. After that you can apply the identities that reduce 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 ${\mathrm{sin}}^{2}x=1-{\mathrm{cos}}^{2}x$. You're left with no trigonometric expressions except $\mathrm{cos}x$. Then the substitution $u=\mathrm{cos}x$ 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 $\mathrm{cos}\left(x+2x\right)$ 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

hence by setting $z=\mathrm{cos}x$ we have to solve:
$4{z}^{3}-4{z}^{2}-3z+3=\left(z-1\right)\left(4{z}^{2}-3\right)=0$
so , from which

nick1337

Using

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