Question

find all solutions to the equation 2cos^2x=13sinx-5 in the interval (0,2pi)

Answer 1

First step is to replace cos^2x with 1 - sin^2x

Answer 2

then waht?

Answer 3

2(1 - sin^2x) = 13sinx - 5
2 - 2sin^2x = 13sinx - 5
0 = 2sin^2x + 13sinx - 7
Now factor it like a quadratic
0 = (2sinx - 1)(sinx + 7)

2sinx - 1 = 0 or sinx + 7 = 0. sinx + 7 can't = 0 since sinx can't = -7..

So that leaves us with 2sinx - 1 = 0
2sinx = 1
sinx = 1/2
x = pi/6 and 5pi/6

Answer 4

oh i see thanks