Find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.) 30 sin2 x = 30 + 15 cos x

Answer 1

angle x = approximately 244.2 degrees + 360n degrees where n = any integer

but 244.2 degrees is the only solution in the domain 0 to 2pi

30sin2x = 30 + 15cosx

use double angle formula

30(2sinxcosx) = 30 + 15cosx

divide by 15

4sinxcosx = 2 + cosx

consider a unit circle where sine = opposite side over hypotenuse, sinx=o/h and

cosine = adjacent side over hypotenuse, cosx=a/h

the hypotenuse = 1, then sinx=o/h = o and cosx=a/h = a

h^2 = a^2 + o^2 = 1^2 = 1

o^2 = 1=a^2

o = sqr(1-a^2)

4sinxcosx = 2 + cosx

4oa = 2 + a

4o = (2+a)/a

o = (2+a)/4a

sqr(1-a^2) = (2+a)/4a

square both sides

1-a^2 = (4+4a + a^2)/16a^2

16a^2 -16a^4 = 4+4a + a^2

16a^4 - 15a^2 +4a + 4 = 0

solve for a

a= -0.435

x = cos^-1(a) = cos^-1(-0.435) = 244.2 degrees

2sin(2(244.2)) = 2 + cos(244.2)

2sin(488.4) = 2 - 0.435

2sin(128.4) = 1.57

2(0.784) = 1.57

1.57 = 1.57