Question

How do you solve `3sin(2)x - sinxcosx - 4cos(2)x = 0` for `-pi<x<pi`?

Answer 1

Solve `f(x) = 3sin 2x - sin x.cos x - 4cos 2x = 0`

Ans: x = 29 and x = 119 deg

Explanation:

Reminder: sin 2x = 2sin x.cos x
`f(x) = 3sin 2x - 1/2(sin 2x) - 4cos 2x = 0`
`f(x) = 5/2 sin 2x - 4cos 2x = 0`
`sin 2x - 8/5 cos 2x = 0`
Put: `tan a = 8/5 = 1.6 = tan 58 = sin 58/cos 58`
`sin 2x.cos 58 - cos 2x.sin 58 = 0`
`sin (2x - 58) = 0` --> `2x - 58 = 0` and `2x - 58 = 180`

a. `sin 2x - 58 = 0` --> `2x = 58`-->`x = 29` deg

b. `sin (2x - 58) = 180` --> `2x = 180 + 58 = 238`--> `x = 119` deg
Check by calculator
a. x = 29 ; sin x = 0.48 ; cos x = 0.87 ; sin 2x = sin 58 = 0.85 ; cos 2x = 0.53 --> f(x) = 3(0.85) - (0.48)(0.87) - 4(0.53) = 0.01 OK
b. x = 119: sin x = 0.87 ; cos x = -0.48 ; sin 2x = sin 238 = -0.85 ; cos 2x = -0.53 --> f(x) = 3(-0.85) - (0.87)(-0.48) + 4(-0.53) = -0.01. OK