Solve 4 * cos(x) * cos(x) - 3 = 0, giving the general solution.

Step-by-step solution

1. 1

   Read the notation. The symbol between the two cosine factors is a multiplication sign, not the variable, so the expression is $4\cos x\cdot\cos x=4\cos^{2}x$ and the equation is

   $4\cos^{2}x-3=0.$

   Writing the repeated factor as a square is the step that makes the equation an ordinary quadratic in $\cos x$.

2. 2

   Isolate $\cos^{2}x$. Adding $3$ and dividing by $4$:

   $\cos^{2}x=\frac34.$

   The value $\tfrac34$ lies in $[0,1]$, so real solutions exist. Had the right side exceeded $1$ there would be none.

3. 3

   Take the square root with both signs. From $\cos^{2}x=\tfrac34$,

   $\cos x=\pm\frac{\sqrt3}{2}.$

   The reference angle for $\tfrac{\sqrt3}{2}$ is $\tfrac{\pi}{6}$ ($30^{\circ}$). Discarding the negative branch would lose half the solutions — a very common slip.

4. 4

   List the base angles in one full turn. For $\cos x=+\tfrac{\sqrt3}{2}$: $x=\tfrac{\pi}{6}$ and $x=\tfrac{11\pi}{6}$. For $\cos x=-\tfrac{\sqrt3}{2}$: $x=\tfrac{5\pi}{6}$ and $x=\tfrac{7\pi}{6}$. Four solutions per period of $2\pi$, symmetric about both axes.

5. 5

   Merge into two compact families. The four angles are $\pm\tfrac{\pi}{6}$ and $\pm\tfrac{\pi}{6}+\pi$, so the general solution is

   $x=\pm\frac{\pi}{6}+k\pi,\qquad k\in\mathbb{Z}.$

   Checking: $4\cos^{2}\!\left(\tfrac{\pi}{6}\right)-3=4\cdot\tfrac34-3=0$ ✓, and the same at $\tfrac{5\pi}{6}$ and $\tfrac{7\pi}{6}$ ✓. The period is $\pi$ rather than $2\pi$ because $\cos^{2}$ is unchanged when $x$ increases by $\pi$.