# Thread: [SOLVED] Another Cos Question. Urgent!!

1. ## [SOLVED] Another Cos Question. Urgent!!

Barron's Math IIC SAT Subject Test Page 82

Which of the following is a solution of cos3x=0.5

a) 60 degrees

b) (5pi)/3

c)cos^-1 (1/6)

d) cos^-1 ( $\frac{\sqrt3}{2}$

e) $\frac{1}{3}$cos^-1(1/2)

2. Originally Posted by fabxx
Barron's Math IIC SAT Subject Test Page 82

Which of the following is a solution of cos3x=0.5

a) 60 degrees

b) $\frac{5\pi}{3}$

c)cos^-1 (1/6)

d) cos^-1 ( $\frac{\sqrt3}{2}$

e) $\frac{1}{3}$cos^-1(1/2)
Solve for x:

$\cos(3x)=\tfrac{1}{2}\implies \cos^{-1}\left[\cos(3x)\right]=\cos^{-1}\left(\tfrac{1}{2}\right)\implies 3x=\cos^{-1}\left(\tfrac{1}{2}\right)$

I'm sure the answer is pretty obvious now

--Chris

3. $cos(3x) = \frac{1}{2}$
$arcos(cos(3x)) = arccos(\frac{1}{2})$
$3x = \frac{\pm\pi}{3}$
$x = \frac{\pm\pi}{9}$

4. Yeah it does thanks but when do you know to use arc-cos or sin? And, if I approach these kind of problems later, do i always use arc-cos?

5. If x is within a cos, you use arccos to get it out of it.
Same thing for sin tan or any trigo function.

6. Originally Posted by fabxx
Yeah it does thanks but when do you know to use arc-cos or sin? And, if I approach these kind of problems later, do i always use arc-cos?
It depends on the problem.

If you had $\sin (x+3)=\tfrac{\sqrt{3}}{2}$, and you wanted to solve for x, you would apply arcsin to both sides, since we know that $\sin^{-1}\left(\sin u\right)=u$

In your problem, you would have to apply arccos to both sides, since we know that $\cos^{-1}\left(\cos u\right)=u$

Does this clarify things?

--Chris