Q : Find all solutions of the equation in the interval [0, 2pi )
2 cos² x = 13 sin x – 5
(Multiple Choice)
A. π/4 , 3π/4 , 5π/4 , 7π/4
B. π/6, 5π/ 6
C. 3π/4 , 7π/4
D. π/3 , 2π/3 , 4π/3 , 5π/3
A: Start with our orignal problem:
2 cos² x = 13 sin x – 5
I don’t want to have sin’s and cos’s mixed, so I need to do some substituting. From a trig identity, we know that:
sin² x + cos² x = 1
Rearrange to get:
cos² x = 1 – sin² x
Substitute that into our equation in place of cos² x like so:
2 cos² x = 13 sin x – 5
2 (1 – sin² x) = 13 sin x – 5
Simplify and rearrange:
2 – 2sin² x = 13 sin x – 5
– 2 sin² x – 13sinx + 7 = 0
Divide everything by -1:
2 sin² x + 13sinx – 7 = 0
Factor:
(2sinx – 1)(sinx + 7) = 0
So, either
(1) 2 sinx – 1 = 0
or
(2) sinx + 7 = 0
Solve (1):
2 sinx – 1 = 0
2 sinx = 1
sin x = 1/2
Refer to our exact values chart to find where sin x = 1/2
We find that sin x = 1/2 when x = π/6… but, we aren’t done because we have to consider ever angle from [0, 2π)
So, we ask ourselves… in what other quadrant is “sin” a positive value (since we have sinx = + 1/2)? This happens in the second quadrant. So, what angle in the second quadrant has a reference angle of π/6 (or 30 degrees?)… To find this, we calculate π – π/6 (which is 180 – 30 in degrees) = 5π/6 (0r 150 degrees).
So, x = π/6, 5π/6
Now solve (2)….
sinx + 7 = 0
sinx = -7
Impossible. No solutions for this part.
So, B is our answer.