# Sin, Cos, Tan

• Feb 13th 2008, 03:50 PM
onenameless
Sin, Cos, Tan
My quesion says:

Solve this equation for 0<x<2(pi)

4sinx cosx -2(square root)3sinx + 2cosx - (square root)3 = 0
• Feb 13th 2008, 05:03 PM
Jhevon
Quote:

Originally Posted by onenameless
My quesion says:

Solve this equation for 0<x<2(pi)

4sinx cosx -2(square root)3sinx + 2cosx - (square root)3 = 0

what you typed is confusing...

i suppose you meant $\displaystyle 4 \sin x \cos x - 2 \sqrt{3} \sin x + 2 \cos x - \sqrt{3} = 0$

in that case, factor out the common $\displaystyle 2 \sin x$ from the first two terms:

$\displaystyle \Rightarrow 2 \sin x ( 2 \cos x - \sqrt{3}) + (2 \cos x - \sqrt{3}) = 0$

but now, $\displaystyle 2 \cos x - \sqrt{3}$ is common to both terms, so pull that out

$\displaystyle \Rightarrow (2 \cos x - \sqrt{3})(2 \sin x + 1) = 0$

now continue...
• Feb 14th 2008, 06:42 PM
onenameless
Yes, that was what i meant, sorry i could not find out how to type in a square root sign. But thanks, that's pretty much how much i needed.
• Feb 14th 2008, 07:10 PM
Jhevon
Quote:

Originally Posted by onenameless
Yes, that was what i meant, sorry i could not find out how to type in a square root sign. But thanks, that's pretty much how much i needed.

so what's the solution? :D
• Feb 15th 2008, 02:20 PM
onenameless
(2sinx + 1)(2cosx-squareroot3) = 0

2sinx +1 = 0
2sinx = -1
sinx = -0.5
x = -30

2cosx - sr(3) = 0
2cosx = sr(3)
cosx = sr(3) / 2
x = 30
• Feb 16th 2008, 04:53 PM
Jhevon
Quote:

Originally Posted by onenameless
(2sinx + 1)(2cosx-squareroot3) = 0

2sinx +1 = 0
2sinx = -1
sinx = -0.5
x = -30

2cosx - sr(3) = 0
2cosx = sr(3)
cosx = sr(3) / 2
x = 30

your solutions are correct, but they are not the solutions we are after. the problem asked for $\displaystyle 0 < x < 2 \pi$, so your first solution is wrong, and your second is incomplete. there are two solutions for each (they coincide), can you find them?
• Feb 16th 2008, 06:09 PM
onenameless
Quote:

Originally Posted by Jhevon
your solutions are correct, but they are not the solutions we are after. the problem asked for $\displaystyle 0 < x < 2 \pi$, so your first solution is wrong, and your second is incomplete. there are two solutions for each (they coincide), can you find them?

1) x = 210 and 330
2) x = 30 and 150
• Feb 16th 2008, 06:11 PM
Jhevon
Quote:

Originally Posted by onenameless
1) x = 210 and 330
2) x = 30 and 150

the one in red is wrong, the others are fine (remember what trig functions are positive in what quadrants)
• Feb 17th 2008, 07:57 AM
onenameless
Quote:

Originally Posted by Jhevon
the one in red is wrong, the others are fine (remember what trig functions are positive in what quadrants)

But sin of x=30 is 0.5, and sin of x=150 is also 0.5, both are the same values with the same positive signs. 30 is in the first quadrant and 150 is in the second, which is where sin is always positive.
• Feb 17th 2008, 07:58 PM
Jhevon
Quote:

Originally Posted by onenameless
But sin of x=30 is 0.5, and sin of x=150 is also 0.5, both are the same values with the same positive signs. 30 is in the first quadrant and 150 is in the second, which is where sin is always positive.

your second equation deals with cosine, not sine
• Feb 18th 2008, 02:38 PM
onenameless
Quote:

Originally Posted by Jhevon
your second equation deals with cosine, not sine

Oh man i did not see that. Right, then it's:
x=30 and 330
• Feb 18th 2008, 02:40 PM
Jhevon
Quote:

Originally Posted by onenameless
Oh man i did not see that. Right, then it's:
x=30 and 330

correct