Solve the inequality 2cos(x) + sqrt(3) >= 0 for x in the interval [-pi, pi].

Step-by-step solution

1. 1

   Isolate the cosine. Subtracting $\sqrt3$ and dividing by the positive number $2$ leaves the inequality direction unchanged:

   $\cos x\ge-\frac{\sqrt3}{2}.$

   Since $-\tfrac{\sqrt3}{2}\approx-0.866$ lies inside $[-1,1]$, the solution set is a genuine sub-interval — neither empty nor everything.

2. 2

   Find the boundary angles. The equation $\cos x=-\tfrac{\sqrt3}{2}$ has reference angle $\tfrac{\pi}{6}$, and cosine is negative in the second and third quadrants, so on $[-\pi,\pi]$ the solutions are

   $x=\frac{5\pi}{6}\quad\text{and}\quad x=-\frac{5\pi}{6}.$

   Cosine is an even function, so its boundary angles on a symmetric interval always come in $\pm$ pairs.

3. 3

   Decide which side of the boundaries to keep. On $[-\pi,\pi]$ the cosine is largest at the centre, $\cos0=1$, and smallest at the two ends, $\cos(\pm\pi)=-1$. So $\cos x\ge-\tfrac{\sqrt3}{2}$ holds on the middle band between the two boundary angles, not outside it:

   $-\frac{5\pi}{6}\le x\le\frac{5\pi}{6}.$

4. 4

   Include the endpoints. The inequality is non-strict, and at $x=\pm\tfrac{5\pi}{6}$ the expression equals $2\left(-\tfrac{\sqrt3}{2}\right)+\sqrt3=0$, which satisfies $\ge0$. Both boundary angles therefore belong to the solution set, and the interval is closed.

5. 5

   Test either side of a boundary. At $x=2.6180\approx\tfrac{5\pi}{6}$: $2\cos x+\sqrt3=0$ ✓ (included). At $x=2.7$, just outside: $2\cos(2.7)+\sqrt3=2(-0.9040721)+1.7320508=-0.0760935<0$ ✓ (excluded), and by symmetry the same happens at $x=-2.7$. At the centre $x=0$: $2+\sqrt3=3.73>0$ ✓.