Homework Help Question & Answers

How do you solve #4sinx-3sin2x=0# for #0<=x<=360#?

How do you solve #4sinx-3sin2x=0# for #0<=x<=360#?
0 0
Next > < Previous
ReportAnswer #1

#0, 48^@19, 180^@, 311^@81, 360^@#

Explanation:

Since sin 2x = 2sin x.cos x, we get:
4sin x - 6sin x.cos x = 0
sin x(4 - 6cos x) = 0
Trig table and unit circle -->
a. sin x = 0 --> #x = 0#, #x = 180^@#, and #x = 360^@#
b. 4 - 6cos x = 0
cos x = 4/6 = 2/3 --> x = +- 48^@19
Arc (-48.19) and arc (360 - 48.19 = 311.81) are co-terminal
Answers for (0, 360)
#0, 48^@19, 180^@, 311^@81, 360^@#

answered by: Nghi N.
Know the answer?
Add Answer of:
How do you solve #4sinx-3sin2x=0# for #0<=x<=360#?
Your Answer: Your Name: What's your source?
Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.