Solve cos(2x - pi/3) + cos(x) = 1/2 for x in [0, 2 pi).

Step-by-step solution

1. 1

   Expand the compound angle. Using $\cos(A-B)=\cos A\cos B+\sin A\sin B$ with $\cos\tfrac\pi3=\tfrac12$, $\sin\tfrac\pi3=\tfrac{\sqrt3}{2}$: $\frac12\cos 2x+\frac{\sqrt3}{2}\sin 2x+\cos x = \frac12.$

2. 2

   Write everything in terms of sin x and cos x. With $\cos 2x = 2c^2-1$ and $\sin 2x = 2sc$ (writing $c=\cos x$, $s=\sin x$): $\frac{2c^2-1}{2}+\sqrt3\,sc+c = \frac12 \quad\Longrightarrow\quad c^2+c-1 = -\sqrt3\,sc.$

3. 3

   Square to eliminate the sine, and note the risk. Replacing $s^2$ by $1-c^2$: $(c^2+c-1)^2 = 3c^2(1-c^2) \quad\Longrightarrow\quad 4c^4+2c^3-4c^2-2c+1 = 0.$ Squaring can create roots that satisfy the quartic but not the original equation, so every candidate must be tested at the end.

4. 4

   Solve the quartic numerically. In $[-1,1]$ the quartic has exactly two roots: $c \approx 0.3369592 \quad\text{and}\quad c \approx 0.8941126.$ Neither is a nice surd, so the answer will be numerical.

5. 5

   Convert each cosine value to angles and filter. Each $c$ gives two angles in $[0,2\pi)$: for $c=0.3369592$, $x=1.227111$ or $5.056074$; for $c=0.8941126$, $x=0.464351$ or $5.818835$. Substituting into the original equation, only $x=1.227111$ and $x=5.818835$ give $\tfrac12$ - the other two are the extraneous roots introduced by squaring (they satisfy the sign-flipped version $c^2+c-1=+\sqrt3 sc$).

6. 6

   Verify the survivors. At $x=1.227111$: $\cos(1.407024)+\cos(1.227111) = 0.163041+0.336959 = 0.500000$. At $x=5.818835$: $\cos(10.590472)+\cos(5.818835) = -0.394113+0.894113 = 0.500000$. Both are correct to the digits shown.