Question

Find the sum of the three smallest positive values of $\theta$ such that $4\cos^2(2\theta-\pi) =3$.

Answer 1

Answer:

$\boxed{\boxed{\dfrac{13\pi}{12}}}$

Step-by-step explanation:

$4\cos^2(2\theta-\pi)=3\qquad\text{divide both sides by 4}\\\\\cos^2(2\theta-\pi)=\dfrac{3}{4}\to\cos(2\theta-\pi)=\pm\sqrt{\dfrac{3}{4}}\\\\\cos(2\theta-\pi)=-\dfrac{\sqrt3}{2}\ \vee\ \cos(2\theta-\pi)=\dfrac{\sqrt3}{2}$

$2\theta-\pi=\dfrac{5\pi}{6}+2k\pi\ \vee\ 2\theta-\pi=-\dfrac{5\pi}{6}+2k\pi\qquad k\in\mathbb{Z}\\\\2\theta-\pi=\dfrac{\pi}{6}+2k\pi\ \vee\ 2\theta-\pi=-\dfrac{\pi}{6}+2k\pi\qquad k\in\mathbb{Z}\\\\\text{add}\ \pi\ \text{to both sides}\\\\2\theta=\dfrac{11\pi}{6}+2k\pi\ \vee\ 2\theta=\dfrac{\pi}{6}+2k\pi\\\\2\theta=\dfrac{7\pi}{6}+2k\pi\ \vee\ 2\theta=\dfrac{5\pi}{6}+2k\pi\\\\\text{divide both sides by 2}$

$\theta=\dfrac{11\pi}{12}+k\pi\ \vee\ \theta=\dfrac{\pi}{12}+k\pi\ \vee\ \theta=\dfrac{7\pi}{12}+k\pi\ \vee\ \theta=\dfrac{5\pi}{12}+k\pi\\\\\qquad k\in\mathbb{Z}$

$\text{The three smallest positive values of}\ \theta:\\\\\dfrac{\pi}{12},\ \dfrac{5\pi}{12},\ \dfrac{7\pi}{12}\\\\\text{The sum:}\\\\S=\dfrac{\pi}{12}+\dfrac{5\pi}{12}+\dfrac{7\pi}{12}=\dfrac{\pi+5\pi+7\pi}{12}=\dfrac{13\pi}{12}$