Solving Trigonometric Functions: $4\cos^2(2\theta-\pi) =3$

Question

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

I have no idea on how to do this problem! Thanks in advance!

Answer 1

HINT

\begin{align*} \cos(2\theta - \pi) = \pm\frac{\sqrt{3}}{2} = \pm\cos\left(\frac{\pi}{6}\right) \end{align*}

Answer 2

Hint:

Use double angle formula

$\cos(x-2\pi)=\cos x$

Alternatively, we can prove $\cos^2y=\cos^2B\implies y=m\pi\pm B$

Where $m$ is any integer

Answer 3

Using method of difference of squares, $$a^2-b^2=(a+b)(a-b)$$

$$4\cos^2(2\theta-\pi)=3$$ $$4\cos^2(2\theta-\pi)-3=0$$ $$(2\cos(2\theta-\pi)+\sqrt3)(2\cos(2\theta-\pi)-\sqrt3)=0$$

We then have two different solution.

$$\cos(2\theta-\pi)=\frac{\sqrt 3}{2}$$

$$\cos(2\theta-\pi)=-\frac{\sqrt 3}{2}$$

Answer 4

$$4\cos^2(2\theta-\pi)=3 \iff 2[\cos (4\theta-2\pi)+1]=3$$

$$\cos (4\theta-2\pi)+1=\frac{3}{2} \iff \cos (4\theta-2\pi)=\frac{1}{2}$$

$$\theta_1=\frac{\pi}{3}$$

$$4\theta-2\pi=\frac{\pi}{3}+2\text n\pi \iff 4\theta=\frac{7\pi}{3}+2\text n\pi$$

$$4\theta=\frac{7\pi}{3}+2\text n\pi \iff \theta=\frac{7\pi}{12}+\frac{\text n\pi}{2}$$

$$\theta_2=2\pi-\frac{\pi}{3}=\frac{5\pi}{3}$$

$$4\theta-2\pi=\frac{5\pi}{3}+2\text n\pi \iff 4\theta=\frac{11\pi}{3}+2\text n\pi$$

$$ 4\theta=\frac{11\pi}{3}+2\text n\pi \iff \theta=\frac{11\pi}{12}+\frac{\text n\pi}{2}$$

where $$ n=0,\pm1,\pm2,...$$

Answer 5

Using $\cos2x = 2\cos^2x-1$, we get $$4\cos^2(2\theta-\pi) = 2[\cos (4\theta-2\pi)+1]= 2\cos 4\theta + 2 = 3$$ Therefore, $$ \cos 4\theta = \frac{1}{2} $$ $$ 4\theta = \pm \left(\frac{\pi}{3} + 2n\pi\right)$$ $$ \theta = \pm\left(\frac{\pi}{12} + \frac{n\pi}{2}\right)$$

It can also be expressed as $$ \theta = \pm \frac{\pi}{12} + \frac{n\pi}{2}$$