# More Trig Solving

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

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.