# Trig cosx question

1. May 14, 2012

### fan_boy17

1. The problem statement, all variables and given/known data

solve 5sinx +12cosx =6.5 between 0 and 180 degrees

2. Relevant equations

3. The attempt at a solution

i tried squaring both sides. (5sinx +12cosx)^2= (6.5)^2

25sinx +60sinxcosx +60sinxcosx + 144cosx^2 =42.25

2. May 14, 2012

### Infinitum

Squaring isn't really a good idea since it might yield extraneous roots. Try writing the given expression as a single trigonometric ratio.

Hint : try converting it into a form of sin(x+T)

Last edited: May 14, 2012
3. May 14, 2012

### HallsofIvy

Staff Emeritus
Sine is always postive between 0 and 180 degrees so you can write that as $5\sqrt{1- cos^2(x)}+ 12cos(x)= 6.5$.

Rewrite it as $5\sqrt{1- cos^2(x)}= 6.5- 12 cos(x)$ and square both sides.

4. May 14, 2012

### Karamata

Or, you can divide equation by $\sqrt{5^2+12^2} = 13$, and you will have
$\frac{5}{13}\sin x + \frac{12}{13}\cos x = \frac{6.5}{13} = \frac{1}{2}$

And you know that $\cos \arccos \frac{5}{13} = \frac{5}{13}$ and $\sin \arccos \frac{5}{13} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \frac{12}{13}$

And then, use formula for $\sin(x+y)$

5. May 14, 2012

### HallsofIvy

Staff Emeritus
Excellent Engliosh, excellent mathematics!

6. May 15, 2012

### Infinitum

This is exactly what I was suggesting.