Question

How do you solve `cos2x+5=4sinx` for `0<=x<=360`?

Answer 1

`pi/2`

Explanation:

cos 2x + 5 = 4sin x
Use trig identity: `cos 2x = 1 - 2sin^2 x`
`1 - 2sin^2 x + 5 = 4sin x`
Bring the equation to standard form:
`2sin^2 x + 4sin x - 6 = 0`
Solve this quadratic equation for sin x.
Since a + b + c = 0, use shortcut. There are 2 real roots:
sin x = 1, and
`sin x = c/a = -6/2 = - 3` (rejected as < -1)
Unit circle gives -->
sin x = 1 --> `x = pi/2`
Answers for `(0, 2pi)`:
`pi/2`
Check.
`x = pi/2` --> sin x = 1 --> `cos 2x = cos pi = - 1`
cos 2x + 5 = 4sin x --> - 1 + 5 = 4. OK