Question

How do you solve the equation for exact solutions in the interval [0,2pi)?

(tanx - 1)(2sinx -1) = 0

Answer 1

`x = pi/6, pi/4, (5pi)/6, (5pi)/4`

Explanation:

The equation has solutions when
`(tanx-1)=0` or `(2sinx-1)=0`
via zero product property

In 0 to `2pi`, we therefore want to find when
`tanx = 1 or sinx = 1/2`

This is easy given our knowledge of the unit circle. Since sine is the `y` coordinate of the unit circle, the angles must be in `[0,pi]`. Thinking about the unit circle, we discover that these are at
`pi/6, (5pi)/6`

We can do the same for tangent, realizing that the angles are
`pi/4, (5pi)/4`

Therefore, all four are zeroes of the above function, hence:
`x = pi/6, pi/4, (5pi)/6, (5pi)/4`