Find the three smallest positive values of $ \theta $ such that $ 4\cos^2(2\theta-\pi) =3. $
I saw that $\cos{(2\theta-\pi)}$ equals $\frac{\sqrt{3}}{2}$, therefore, $2\theta-\pi=\frac{\pi}{6}+2x\pi$ or $2\theta-\pi=\frac{11\pi}{6}+2x\pi.$
This gives the three smallest values as $\frac{5\pi}{12}$, $\frac{7\pi}{12}$, and $\frac{17\pi}{12}$. However, these answers are not correct, so where did I go wrong?
Whenever we have squares of any Trigonometric ratio, we can convert it to respective double angle of cosine
See also : Writing answers to trigonometric equation
$\displaystyle 4\cos^2(2\theta-\pi) =3\implies\cos\{2(2\theta-\pi)\}=2\cos^2(2\theta-\pi)-1=\frac32-1=\frac12$
$\displaystyle\implies2(2\theta-\pi)=2n\pi\pm\frac\pi3$ where $n$ is any integer
$\displaystyle\implies\theta=\frac{n\pi}2+\pi\pm\frac\pi{12}$
Taking '+' sign, we need $\displaystyle 0<\frac{n\pi}2+\pi+\frac\pi{12}<2\pi$
Multiplying each by $\dfrac{12}\pi,$
$\displaystyle\implies 0<6n+12+1<24\iff -13<6n<11\iff -2\le n\le1$ as $n$ is any integer
Similarly, for the '-' sign
Your equation is
$$ 4\cos^2(2\theta-\pi)=3. $$
When taking the square root, do not forget to account for the nonpositive roots. More specifically, the solutions to the equation are
$$\cos(2\theta-\pi)=\frac{\sqrt{3}}{2},\text{ or }\cos(2\theta-\pi)=-\frac{\sqrt{3}}{2},$$
which proves that the following $\theta$ are the solutions of the equation:
$$\theta=\frac{7\pi}{12}+k\pi,\text{ or }\theta=\frac{5\pi}{12}+k\pi,\text{ or }\theta=\frac{\pi}{12}+k\pi,\text{ or }\theta=\frac{11\pi}{12}+k\pi,$$
where in these equations $k\in\mathbb{Z}$ is arbitrary.
Thus the smallest positive $\theta$ solving the equation are $\theta_1=\frac{\pi}{12}$, $\theta_2=\frac{5\pi}{12}$, $\theta_3=\frac{7\pi}{12}$.
You seem to have forgotten to divide by 2 when solving for $\theta$.
– Ian May 03 '14 at 02:03