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

1 Answer
Nghi N.
Aug 21, 2015

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

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
3537 views around the world
You can reuse this answer
Creative Commons License