Question

How do you solve `7sinx+5=2cos^2x` in the interval `0<=x<=2pi`?

Answer 1

`(7pi)/6 and (11pi)/6`

Explanation:

`7sin x + 5 = 2cos^2 x.`
Replace `cos^2 x` by `(1 - sin^2 x)`:
`7sin x + 5 = 2 - 2sin^2 x`
`2sin^2 x + 7sin x + 3 = 0`
Solve this quadratic equation for sin x.
`D = d^2 = b^2 - 4ac = 49 - 24 = 25` --> `d = +- 5`
There are 2 real roots:
`x = -b/(2a) = -7/4 +- 5/4 = (-7 +- 5)/4`
`x1 = -2/4 = -1/2`, and `x2 = -12/4 = -3` (Rejected because < -1)
Trig table, and unit circle give -->
`x = - 1/2` --> 2 arcs x
a. `x = - pi/6`, or `x = (11pi)/6`--> (co-terminal), and
b. `x = pi + pi/6 = (7pi)/6`
Answers for `(0, 2pi)`
`(7pi)/6 and (11pi)/6`