How to solve | 2 cos x + 3 sin x | = sin x?

Question

How do you particularly solve the question with | | sign?

Answer 1

To solve an equation |f(x)| = g(x), we split this up into two equations:
f(x) = g(x), and f(x) = -g(x). Therefore,

|2cos(x) + 3sin(x)| = sin(x) gets split into

1) 2cos(x) + 3sin(x) = sin(x), and
2) 2cos(x) + 3sin(x) = -sin(x).

Let's solve these separately. I'm going to assume there is a restriction of 0 <= x < 2pi.

1) 2cos(x) + 3sin(x) = sin(x)

Move 3sin(x) to the right hand side to obtain

2cos(x) = -2sin(x)

Divide both sides by 2cos(x),

1 = -sin(x)/cos(x). By definition, this is
1 = -tan(x), so
-1 = tan(x).

tan(x) = -1 at the points 3pi/4 and 7pi/4. Therefore,
x = {3pi/4, 7pi/4}

2) 2cos(x) + 3sin(x) = -sin(x)

Move 3sin(x) over to the right hand side,

2cos(x) = -4sin(x)

Divide both sides by 2cos(x),

1 = -2tan(x), so

-1/2 = tan(x)

Unfortunately, -1/2 isn't one of our known values on the unit circle. we can, however, take the arctan of both sides to obtain
x = arctan(-1/2). Since the solutions of tan are "pi" away from each other (the same way 3pi/4 and 7pi/4 are "pi" distance apart), our solutions are

x = {arctan(-1/2), arctan(-1/2) + pi}

Combining the results from both equations,

x = {3pi/4, 7pi/4, arctan(-1/2), arctan(-1/2) + pi}

Answer 2

This is really two problems:

1) 2 cos x + 3 sin x = sin x

and

2) -(2 cos x + 3 sin x) = sin x

Solve both to find the solutions

Answer 3

Treat is as two equations:
|2cosx + 3sinx | = sinx
→ 2cosx + 3sinx = sinx OR -cosx - 3sinx = sinx
→ 2cosx = -2sinx OR -cosx = 4sinx
→ tanx = -1 OR tanx = -1/4
→ x = 3pi/4 + npi OR x = arctan(-1/4) + npi

Answer 4

Yes not easy