Notice that the left side of the equation is the difference formula for cosine.

[latex]\begin{align} \cos x\cos \left(2x\right)+\sin x\sin \left(2x\right)&=\frac{\sqrt{3}}{2} \\ \cos \left(x - 2x\right)&=\frac{\sqrt{3}}{2}&& \text{Difference formula for cosine} \\ \cos \left(-x\right)&=\frac{\sqrt{3}}{2}&& \text{Use the negative angle identity}. \\ \cos x&=\frac{\sqrt{3}}{2}\end{align}[/latex]

From the unit circle in Sum and Difference Identities, we see that [latex]\cos x=\frac{\sqrt{3}}{2}[/latex] when [latex]x=\frac{\pi }{6},\frac{11\pi }{6}[/latex].

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

[latex]\begin{gathered}\cos \left(2\theta \right)=\cos \theta \\ 2{\cos }^{2}\theta -1=\cos \theta \\ 2{\cos }^{2}\theta -\cos \theta -1=0 \\ \left(2\cos \theta +1\right)\left(\cos \theta -1\right)=0 \\ 2\cos \theta +1=0 \\ \cos \theta =-\frac{1}{2} \\ \text{ } \\ \cos \theta -1=0 \\ \cos \theta =1 \end{gathered}[/latex]

So, if [latex]\cos \theta =-\frac{1}{2}[/latex], then [latex]\theta =\frac{2\pi }{3}\pm 2\pi k[/latex] and [latex]\theta =\frac{4\pi }{3}\pm 2\pi k[/latex]; if [latex]\cos \theta =1[/latex], then [latex]\theta =0\pm 2\pi k[/latex].

If we rewrite the right side, we can write the equation in terms of cosine:

[latex]\begin{align} 3 cos\theta +3& ={2 sin}^{2}\theta \\ 3 cos\theta +3& =2\left(1-{\text{cos}}^{2}\theta \right) \\ 3 cos\theta +3& =2 - 2{\cos }^{2}\theta \\ 2{\cos }^{2}\theta +3 cos\theta +1& =0 \\ \left(2 cos\theta +1\right)\left(\cos \theta +1\right)& =0 \\ 2 cos\theta +1& =0 \\ \cos \theta & =-\frac{1}{2} \\ \theta & =\frac{2\pi }{3},\frac{4\pi }{3} \\ \text{ } \\ \cos \theta +1& =0 \\ \cos \theta & =-1 \\ \theta & =\pi\end{align}[/latex]

Our solutions are [latex]\theta =\frac{2\pi }{3},\frac{4\pi }{3},\pi[/latex].

We can see that this equation is the standard equation with a multiple of an angle. If [latex]\cos \left(\alpha \right)=\frac{1}{2}[/latex], we know [latex]\alpha[/latex] is in quadrants I and IV. While [latex]\theta ={\cos }^{-1}\frac{1}{2}[/latex] will only yield solutions in quadrants I and II, we recognize that the solutions to the equation [latex]\cos \theta =\frac{1}{2}[/latex] will be in quadrants I and IV.

Therefore, the possible angles are [latex]\theta =\frac{\pi }{3}[/latex] and [latex]\theta =\frac{5\pi }{3}[/latex]. So, [latex]2x=\frac{\pi }{3}[/latex] or [latex]2x=\frac{5\pi }{3}[/latex], which means that [latex]x=\frac{\pi }{6}[/latex] or [latex]x=\frac{5\pi }{6}[/latex]. Does this make sense? Yes, because [latex]\cos \left(2\left(\frac{\pi }{6}\right)\right)=\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}[/latex].

Are there any other possible answers? Let us return to our first step.

In quadrant I, [latex]2x=\frac{\pi }{3}[/latex], so [latex]x=\frac{\pi }{6}[/latex] as noted. Let us revolve around the circle again:

[latex]\begin{align} 2x&=\frac{\pi }{3}+2\pi \\ &=\frac{\pi }{3}+\frac{6\pi }{3} \\ &=\frac{7\pi }{3} \end{align}[/latex]

so [latex]x=\frac{7\pi }{6}[/latex].

One more rotation yields

[latex]\begin{align} 2x&=\frac{\pi }{3}+4\pi \\ &=\frac{\pi }{3}+\frac{12\pi }{3} \\ &=\frac{13\pi }{3} \end{align}[/latex]

[latex]x=\frac{13\pi }{6}>2\pi[/latex], so this value for [latex]x[/latex] is larger than [latex]2\pi[/latex], so it is not a solution on [latex]\left[0,2\pi \right)[/latex].

In quadrant IV, [latex]2x=\frac{5\pi }{3}[/latex], so [latex]x=\frac{5\pi }{6}[/latex] as noted. Let us revolve around the circle again:

[latex]\begin{align}2x&=\frac{5\pi }{3}+2\pi \\ &=\frac{5\pi }{3}+\frac{6\pi }{3} \\ &=\frac{11\pi }{3}\end{align}[/latex]

so [latex]x=\frac{11\pi }{6}[/latex].

One more rotation yields

[latex]\begin{align}2x&=\frac{5\pi }{3}+4\pi \\ &=\frac{5\pi }{3}+\frac{12\pi }{3} \\ &=\frac{17\pi }{3} \end{align}[/latex]

[latex]x=\frac{17\pi }{6}>2\pi[/latex], so this value for [latex]x[/latex] is larger than [latex]2\pi[/latex], so it is not a solution on [latex]\left[0,2\pi \right)[/latex].

Our solutions are [latex]x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\text{and }\frac{11\pi }{6}[/latex]. Note that whenever we solve a problem in the form of [latex]\sin \left(nx\right)=c[/latex], we must go around the unit circle [latex]n[/latex] times.