Solve Simple Trig Question: Sin x = √3 * Cos X

[kscplay]

Homework Statement

-sin x = √3 * cos x
where x is [0,2π]

Homework Equations

The Attempt at a Solution

Would it be wrong to square both sides and then factor?
sin² x = 3cos² x
1-cos² x = 3cos² x
0= 4cos² x - 1
0 = (2cos x -1)(2cos x + 1)
Now I solve the factors (2 solutions for each)

Since I've squared both sides, haven't I introduced extra solutions that are not part of the original problem? Will all 4 resulting solutions be correct? Thanks.

 

[Shootertrex]

Well, when you square root something, you will get a positive and a negative answer, which you already know. Since you defined x as [0,2π], there are no negative numbers in that interval. Therefore, you may disregard the negative answers for this interval, which should leave you with two answers.

 

[tiny-tim]

hi kscplay!

-sin x = √3 * cos x
where x is [0,2π]
…
Since I've squared both sides, haven't I introduced extra solutions that are not part of the original problem?

yes, since you will also be finding the solutions to sin x = √3 * cos x

the same x can't be a solution to both (unless sinx = 0, of course)

i think the only way to decide which solutions to reject is to actually check each one ​

 

[HallsofIvy]

But the simplest way to solve that equation is to divide both sides by -cos(x):
\frac{sin(x)}{cos(x)}= tan(x)= -\sqrt{3}