solve sin(3θ) = cos(2θ) for 0 ≤ θ ≤ 2π

Answer 1

For the sake of simplifying the notation, I omit the angle theta after everything

becomes sines and cosine

0 =

sin(3T) - cos(2T) = <--- everything on one side, zer0 on the other

sin(2T+t) - cos(2t) =

sin(2T)cost + cos(2t)sin t - cos(2t) = <--- angle addition formula

2 sin cos^2 + (2 cos^2 - 1) sin - cos 2t = <--- double angle formulas

sin ( 2 cos^2 + 2 cos^2 - 1) - cos 2t = <--- factors out sin

sin ( 4 cos^2 - 1) - cos 2t = <--- combines like terms

sin ( 4 cos^2 - 1) - (1 - 2 sin^2) = <-- double angle formula

sin ( 4 ( 1-sin^2) -1) - (1 - 2 sin^2)= <--- trig identity sin^2 + cos^2 = 1 , so cos^2 = 1 - sin^2

sin ( 4 - 4sin^2 - 1) - (1 - 2 sin^2) = <--- distributive

4 sin - 4 sin^3 - sin - 1 + 2 sin^2 = <--- distriburive again

3 sin - 4 sin^3 - 1 + 2 sin^2 <--- combines like terms again

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changing the signs

4 sin^3 - 2 sin^2 - 3 sin + 1 = 0

Let z = sin t

4 z^3 - 2z^2 - 3z + 1 = 0

Rational root theorem is not needed as x=1 is a solution. (Plug in z=1... it works!)

Synthetic division says:

1 | 4 -2 -3 1

________ 4_ 2_ -1_____________

4 2 -1 0

(z-1)(4z^2 + 2z - 1) = 0

quadratic formula says:

z = [-2 +or- sqrt ( 4 - 4(4)(-1))] / 8

z = [ -2 +or- sqrt ( 4 + 16)]/8

= [ -2 +or- sqrt( 20)] / 8

= [ -2 +or- 2sqrt(5)]/8

= [-1 +or- sqrt(5)]/4

FOr z = 1 = sin t , then t = pi/2

t = inverse-sin(z) = arccos(z)

For z = (-1 + sqrt(5))/4 ---> t = 18 degrees = pi/10

For z = (-1 - sqrt(5))/4 ---> t = -54 degrees = 306 degrees = 17*pi/10