Question

How to determine the number of solutions for trig equations? for example: (a sinx - b)(a cosx -a)(b sinx +a)=0; 0 14 years ago

Answer 1

Since we have:\[(a\sin x - b)(a\cos x -a)(b\sin x+a)=0\]each term could be zero. That gives three equations:\[a\sin x - b=0\Longrightarrow \sin x = \frac{b}{a}\]\[a\cos x -a=0\Longrightarrow \cos x = 1\Longrightarrow x=0\]\[b\sin x +a = 0\Longrightarrow \sin x = -\frac{a}{b}\]

Answer 2

The middle term only produces 1 solution, while the other two terms can produce 1 or 2 solutions, depending on what a and b are. So you can have a max of 5 solutions.

Answer 3

why can the first and last terms produce 1 OR 2 solutions?

Answer 4

Think of sin as the y axis of the unit circle. For most values of sin, there are two angles that will have give that value. For example:\[\sin \theta =\frac{1}{2}\Longrightarrow \theta = \frac{\pi}{6}, \theta = \frac{5\pi}{6}\]The only values of sin that have one solution are:\[\sin \theta = 1,\sin \theta = -1\]

Answer 5

OHH! so it's not because of the variables (a,b,c) but the sin and cos in the terms?!

Answer 6

yes :) its just a property that sine and cosine have.

Answer 7

but it says that the answer is only 3 solutions?

Answer 8

Ohh!! it says 0<a<b so we can eliminate the two "other" possible sine solutions that are -1!

Answer 9

oh i didnt see that! if 0<a<b, then \[\frac{b}{a}>1\], so that first equation doesnt have any solutions since sine is always between -1 and 1..

Answer 10

=) thats ok!! Thank you so much!