Question

Verify that the given value is a solution for the trigonometric equation by showing that it makes the equation true:

1. 2 cos^2 x-cos x=0; 2π/3


2. 2 cos^2 x-cos x=0; 0

Answer 1

`\bf \begin{array}{llll} 2cos^2(x)-cos(x)&=&0\\\\ 2cos^2\left( (2\pi )/(3) \right)-cos\left( (2\pi )/(3) \right)\\\\ 2\left[ cos\left( (2\pi )/(3) \right) \right]^2 -cos\left( (2\pi )/(3) \right)\\\\ 2\left[ -(1)/(2) \right]^2-\left( -(1)/(2) \right)\\\\ 2\left[ \cfrac{1}{4} \right]+\cfrac{1}{2}\\\\ \cfrac{1}{2}+\cfrac{1}{2}\\\\ 1&\\e & 0 \end{array} \\\\[-0.35em] ~\dotfill`

`\bf \begin{array}{llll} 2cos^2(x)-cos(x)&=&0\\\\ 2cos^2(0)-cos(0)\\\\ 2[cos(0)]^2-cos(0)\\\\ 2[1]^2-1\\\\ 2-1\\\\ 1&\\e & 0 \end{array}`