Question

\cos ( 2x- \dfrac{ 3 \pi }{ 4 } ) = - \dfrac{ 1 }{ 2 }

Answer 1

Answer:

$\displaystyle{x= \dfrac{17\pi}{24}+ \pi n, \dfrac{25\pi}{24}+\pi n \ \ \ (n \in \mathbb{I})}$

Step-by-step explanation:

Given the cosine equation:

$\displaystyle{\cos \left(2x - \dfrac{3\pi}{4}\right) = -\dfrac{1}{2}}$

Let:

$\displaystyle{\theta = 2x - \dfrac{3\pi}{4}}$

Then we will have:

$\displaystyle{\cos \theta = -\dfrac{1}{2}}$

We know that cosθ is negative in second quadrant and third quadrant.

In second quadrant:

$\displaystyle{\pi - \dfrac{\pi}{3}}\\\\\displaystyle{=\dfrac{3\pi}{3} - \dfrac{\pi}{3}}\\\\\displaystyle{=\dfrac{2\pi}{3}}$

In third quadrant:

$\displaystyle{\pi + \dfrac{\pi}{3}}\\\\\displaystyle{=\dfrac{3\pi}{3} + \dfrac{\pi}{3}}\\\\\displaystyle{=\dfrac{4\pi}{3}}$

Therefore, we will have two solutions where:

$\displaystyle{\theta_1 = \dfrac{2\pi}{3}}\\\displaystyle{\theta_2 = \dfrac{4\pi}{3}}$

Convert from theta back to the original expression:

$\displaystyle{2x_1-\dfrac{3\pi}{4}= \dfrac{2\pi}{3}}\\\\\displaystyle{2x_2-\dfrac{3\pi}{4}= \dfrac{4\pi}{3}}$

Convert two equations into one:

$\displaystyle{2x-\dfrac{3\pi}{4}= \dfrac{2\pi}{3}, \dfrac{4\pi}{3}}$

Since you do not specify the interval, add $\displaystyle{2\pi n}$ where $\displaystyle{n \in \mathbb{I}}$.

$\displaystyle{2x-\dfrac{3\pi}{4}= \dfrac{2\pi}{3}+2\pi n, \dfrac{4\pi}{3}+2\pi n}$

Solve the equation for x:

$\displaystyle{2x= \dfrac{2\pi}{3}+\dfrac{3\pi}{4}+2\pi n, \dfrac{4\pi}{3}+\dfrac{3\pi}{4}+2\pi n}\\\\\displaystyle{2x= \dfrac{8\pi}{12}+\dfrac{9\pi}{12}+2\pi n, \dfrac{16\pi}{12}+\dfrac{9\pi}{12}+2\pi n}\\\\\displaystyle{2x= \dfrac{17\pi}{12}+2\pi n, \dfrac{25\pi}{12}+2\pi n}\\\\\displaystyle{x= \dfrac{17\pi}{24}+ \pi n, \dfrac{25\pi}{24}+\pi n}$

Therefore, the solution is:

$\displaystyle{x= \dfrac{17\pi}{24}+ \pi n, \dfrac{25\pi}{24}+\pi n \ \ \ (n \in \mathbb{I})}$