Question

Solve the equation on the interval [0, 2pi): 12cos^2 x-9=0 What to do?

Answer 1

Answer:

$\large\boxed{x=\dfrac{\pi}{6}\ \vee\ x=\dfrac{5\pi}{6}\ \vee\ x=\dfrac{7\pi}{6}\ \vee\ x=\dfrac{11\pi}{6}}$

Step-by-step explanation:

$12\cos^2x-9=0\qquad\text{add 9 to both sides}\\\\12\cos^2x=9\qquad\text{divide both sides by 12}\\\\\cos^2x=\dfrac{9}{12}\\\\\cos^2x=\dfrac{9:3}{12:3}\\\\\cos^2x=\dfrac{3}{4}\to\cos x=\pm\sqrt{\dfrac{3}{4}}\\\\\cos x=\pm\dfrac{\sqrt3}{2}\\\\\text{Look at the picture}\\\\x=\dfrac{\pi}{6}\ \vee\ x=\dfrac{5\pi}{6}\ \vee\ x=\dfrac{7\pi}{6}\ \vee\ x=\dfrac{11\pi}{6}$