Solve 5sinx +12cosx=6.5: 0-180 Degrees

[fan_boy17]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

 

[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)

 

[HallsofIvy]

Sine is always positive 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.

 

[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)

sorry for bad English

 

[HallsofIvy]

Excellent Engliosh, excellent mathematics!

 

[Infinitum]

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)

sorry for bad English

This is exactly what I was suggesting.