Question

How do you solve `1+sinx=2cos^2x` in the interval `0<=x<=2pi`?

Answer 1

Based on two different cases: `x = pi/6, (5pi)/6 or (3pi)/2`

Look below for the explanation of these two cases.

Explanation:

Since, `cos^ x + sin^2 x = 1`
we have: `cos^2 x = 1 - sin^2 x`

So we can replace `cos^2 x` in the equation `1 + sinx = 2cos^2x` by `(1- sin^2 x)`

`=>2 (1 - sin^2 x) = sin x +1`

or, `2 - 2 sin^2 x = sin x + 1`

or, `0=2sin ^2 x + sin x + 1 - 2`

or, `2sin ^2 x + sin x - 1 = 0`

using the quadratic formula:

`x = (-b+-sqrt(b^2 - 4ac))/(2a)` for quadratic equation `ax^2+bx+c=0`

we have:

`sin x = (-1+-sqrt(1^2 - 4*2*(-1)))/(2*2)`

or, `sin x = (-1+-sqrt(1 + 8))/4`

or, `sin x = (-1+-sqrt(9))/4`

or, `sin x = (-1+-3)/4`

or, `sin x = (-1+3)/4, (-1-3)/4`

or, `sin x = 1/2, -1`

Case I:

`sin x = 1/2`

for the condition: `0<=x<=2pi`

we have:

`x= pi/6 or (5pi)/6` to get positive value of `sinx`

Case II:

`sin x = -1`

we have:

`x= (3pi)/2` to get negative value of `sinx`