Question

Find all solutions in the interval [0,2pi) of the equation: 2cos(4x)-1=0

Answer 1

`\bf 2cos(4x)-1=0\implies 2cos(4x)=1\implies cos(4x)=\cfrac{1}{2} \\\\\\ cos^(-1)\left[ cos(4x) \right]=cos^(-1)\left[ \cfrac{1}{2} \right]\implies 4x=cos^(-1)\left( \cfrac{1}{2} \right) \\\\\\ \measuredangle x = \cfrac{cos^(-1)\left( (1)/(2) \right)}{4}`

check your Unit Circle for all angles on that range that have a cosine of 1/2
then divide that by 4... for example hmmmm say
`\bf (\pi )/(3)` is one, thus x =
`\bf \cfrac{(\pi )/(3)}{4}\implies \cfrac{\pi }{3}\cdot \cfrac{1}{4}\implies \cfrac{\pi }{12}`

that's one