8sin(2x)+13cos(x)=0

Question

Solve the equation for all solutions in the interval

Answer 1

8sin(2x) - 13cosx = 0

8(2sinxcosx) - 13cosx = 0

cosx(16sinx - 13) = 0

cosx = 0 or sinx = 13/16

If cosx = 0, then x = π/2 or 3π/2

If sinx = 13/16, then x = Sin^-1(13/16) or π - Sin^-1(13/16) = 0.95 or 2.19

Solutions in [0, 2π): π/2, 3π/2, 0.95, 2.19

Answer 2

8 ( 2 sin x cos x ) + 13 cos x = 0

16 sin x cos x + 13 cos x = 0

16 sqrt(1 - cos^2) cos + 13 cos = 0

Let T = cos

16 sqrt(1 - T^2) T + 13 T = 0

16 sqrt( 1 - T^2) T = -13 T

16 sqrt(1 - T^2) = -13

sqrt(1 - T^2) = -13/16

1 - T^2 = 169/256

-T^2 = 169/256 - 1

-T^2 = -87/256

T^2 = 87/256

T = +-sqrt(87)/16

cos x = +- sqrt(87)/16

x = arccos( +- sqrt(87)/16)

54.341-

125.6591-

234.341-

305.6591-