Question

How do you solve `cos x = sin 2x`, within the interval [0,2pi)?

Answer 1

`pi/6 , pi/2 , (5pi)/6 , (3pi)/2`

Explanation:

Using the `color(blue)" double angle formula "`

sin2A = 2sinAcosA

hence: cosx = 2sinxcosx

and 2sinxcosx - cosx = 0

take out the common factor cosx

cosx( 2sinx - 1 ) = 0

We now have : cosx = 0 or 2sinx - 1 = 0 → `sinx = 1/2`

solve : cosx = 0 `rArr x = pi/2 , x = (3pi)/2`

solve `sinx = 1/2 rArr x = pi/6 , x = (5pi)/6`

In the interval `[ 0 , 2pi )`

solutions are `color(red)(|bar(ul(color(white)(a/a)color(black)(pi/6 ,pi/2 , (5pi)/6 , (3pi)/2)color(white)(a/a)|)))`