Solve cos(2x - pi/3) + cos x = 0.5 for x in [0, 2pi).

Step-by-step solution

1. 1

   Combine the two cosines with the sum-to-product identity. Using $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ with $A=2x-\frac{\pi}{3}$ and $B=x$: $\frac{A+B}{2}=\frac{3x}{2}-\frac{\pi}{6},\qquad \frac{A-B}{2}=\frac{x}{2}-\frac{\pi}{6},$ so the equation becomes $2\cos\!\left(\frac{3x}{2}-\frac{\pi}{6}\right)\cos\!\left(\frac{x}{2}-\frac{\pi}{6}\right)=\frac12,\quad\text{i.e.}\quad \cos\!\left(\frac{3x}{2}-\frac{\pi}{6}\right)\cos\!\left(\frac{x}{2}-\frac{\pi}{6}\right)=\frac14.$

2. 2

   Recognise that no closed form exists. The two cosine arguments have frequencies in the ratio $3:1$, so the product cannot be reduced to a single trigonometric function of one angle. Substituting $t=\tan\frac{x}{2}$ turns the equation into a quartic in $t$ with irrational roots, so a numerical solution is the honest route here.

3. 3

   Bound the number of solutions. Expanding with the double-angle formula, $\cos(2x-\tfrac{\pi}{3})=\tfrac12\cos 2x+\tfrac{\sqrt3}{2}\sin 2x$, so the left-hand side is a smooth $2\pi$-periodic combination of the first two harmonics. Sampling it over $[0,2\pi)$ shows it runs between $-1.583246$ and $1.891702$ and crosses the level $0.5$ exactly twice, so we expect two solutions.

4. 4

   Bracket and bisect the two crossings. Set $f(x)=\cos(2x-\tfrac{\pi}{3})+\cos x-0.5$. Then $f(1.2)=+0.078629$ and $f(1.3)=-0.214508$ bracket the first crossing, while $f(5.8)=-0.042926$ and $f(5.9)=+0.187084$ bracket the second. Bisecting each bracket to full double precision yields $x_{1}=1.227110981,\qquad x_{2}=5.818834845.$

5. 5

   Substitute both back. At $x_{1}=1.227110981$: $\cos(2x_{1}-\tfrac{\pi}{3})=0.163041$ and $\cos x_{1}=0.336959$, and $0.163041+0.336959=0.500000$. At $x_{2}=5.818834845$: $\cos(2x_{2}-\tfrac{\pi}{3})=-0.394113$ and $\cos x_{2}=0.894113$, and $-0.394113+0.894113=0.500000$. Both solutions check to six decimals.

6. 6

   Convert to degrees and note the general solution. $x_{1}\approx 70.3083^\circ$ and $x_{2}\approx 333.3947^\circ$. Because the whole equation has period $2\pi$, every solution is $x\approx 1.227111+2k\pi\quad\text{or}\quad x\approx 5.818835+2k\pi,\qquad k\in\mathbb{Z}.$