Question

(Solving trig equations)

"Find all solutions of: sin^2(x) + 2sin(x)cos(x) - 3cos^2(x) = 0. Hint: Express in terms of tan(x)."

Answer 1

I don't feel like I'm on the right track with this:

Answer 2

@pgpilot326 :)

Answer 3

factor...
\[\sin^2x+2\sin x\cos x-3\cos^2 x=0\Rightarrow (\sin x-\cos x)(\sin x + 3\cos x)=0\]

set each factor = to 0 and solve.

Answer 4

that's when you'll use the tangent hint.

Answer 5

Thanks for your reply, I'll give it a go.

Answer 6

have fun... gotta run now. don't forget that tangent has a period of pi so add integer multiples of pi to your solutions.

Answer 7

Thanks for your help!

Answer 8

I'm struggling to figure out how to write it in terms of tan(x).

Answer 9

Is this the right thing to do?

Answer 10

\[\sin x - \cos x = 0 \Rightarrow \sin x = \cos x \Rightarrow \frac{\sin x}{\cos x} = 1 \Rightarrow \tan x = 1\]

Answer 11

Sweet! I was on the right track with that. I think I've got it now, I'll keep going.

Answer 12

so what values of x make tan x = 1?

Answer 13

remember the bit about integer multiples of pi...

Answer 14

I'm not sure if this is right.

Answer 15

I've noticed you've been in this thread for a while @ranga , I'm not sure if you're AFK or not, but did I do it right?

Answer 16

sorry. I was answering a couple of other questions and haven't even read this page yet. Are you still looking for an answer?

Answer 17

Oh, no problem! Yeah, in my second last post I attached an image with my answer but I'm not sure if it's right.

Answer 18

2sin(x) - 3cos(x) = -sin(x)tan(x)
divide by -sin(x)
-2 + 3cos(x)/sin(x) = tan(x)
-2 + 3/tan(x) = tan(x)
multiply by tan(x)
-2tan(x) + 3 = tan^2(x)
tan^2(x) + 2tan(x) - 3 = 0
This is a quadratic equation in tan(x).
Solve for tan(x) and then x.

Answer 19

I was starting from the 4th line in your answer.

Answer 20

Ohh, thank you! I'll give it another go.

Answer 21

alright.

Answer 22

BTW, we can get to the quadratic equation in tan(x) from the given problem quickly as follows:
Given problem: sin^2(x) + 2sin(x)cos(x) - 3cos^2(x) = 0
divide throughout by cos^2(x) (assuming cos^2(x) is not zero. When you finally solve for x you should exclude those x that makes cos^2(x) zero).
tan^2(x) + 2tan(x) - 3 = 0
solve the quadratic in tan(x) and then x.

Answer 23

tan^2(x) + 2tan(x) - 3 = 0
tan^2(x) + 3tan(x) - tan(x) - 3 = 0
tan(x)(tan(x) + 3) - 1(tan(x) + 3) = 0
(tan(x) - 1)(tan(x) + 3) = 0
tan(x) = 1 or tan(x) = -3
x = arctan(1) or arctan(-3)

Answer 24

Ohh, interesting. So I did it two different ways:

What's the significance of the negative signs?

Answer 25

I will just do the positive part and since trigonometric functions are periodic you can write the solution as (solutions within 2pi) plus/minus 2n(pi) where n = 0, 1, 2, etc.

Answer 26

Since tan has a period of pi, shouldn't the solutions be +- (pi)n?

Answer 27

Thanks for your help!

Answer 28

you are welcome.