Trigonemetric Functions - Need help understanding constraints

**l flipboi l** · November 6th 2009, 10:42 PM

Hello,

Can someone please help me understand how the answers were obtained?

1. (sqrt3)tanx+1=0 -----> tanx = -1/(sqrt3) or 30 deg.

answer is: x = 5pi/6 + kpi

2. 4(cos^2)x-1=0 -----> cosx = 1/2 or 60 deg.

answer is: x = pi/3 + kpi, 2pi/3 + kpi

3. 3(csc^2)x-4=0 -----> cscx=2/(sqrt3) or 60 deg.

answer is: x = pi/3 + kpi, 2pi/3 + kpi

.................

I can do all the algebraic steps, what I'm confused about is how did they come up with kpi as the answer shouldn't it be 2kpi?

because this is what i got...

1. [5pi/6 + 2kpi, 11pi/6 + 2kpi]
what I was thinking was that since tanx=-1/(sqrt3) I would need the solutions of the quadrant where tan is negative, which is quadrant II and IV

2. [pi/3 + 2kpi, 5pi/3 + 2kpi]
again, since cosx=1/2, which is positive...the solutions would be in quadrant I and IV

3...same idea, positive answer cscx = 2/(sqrt3)..solutions are in quadrant I and II

Please help, thanks!

**Prove It** · November 6th 2009, 11:07 PM

Originally Posted by l flipboi l

Hello,

Can someone please help me understand how the answers were obtained?

1. (sqrt3)tanx+1=0 -----> tanx = -1/(sqrt3) or 30 deg.

answer is: x = 5pi/6 + kpi

2. 4(cos^2)x-1=0 -----> cosx = 1/2 or 60 deg.

answer is: x = pi/3 + kpi, 2pi/3 + kpi

3. 3(csc^2)x-4=0 -----> cscx=2/(sqrt3) or 60 deg.

answer is: x = pi/3 + kpi, 2pi/3 + kpi

.................

I can do all the algebraic steps, what I'm confused about is how did they come up with kpi as the answer shouldn't it be 2kpi?

because this is what i got...

1. [5pi/6 + 2kpi, 11pi/6 + 2kpi]
what I was thinking was that since tanx=-1/(sqrt3) I would need the solutions of the quadrant where tan is negative, which is quadrant II and IV

2. [pi/3 + 2kpi, 5pi/3 + 2kpi]
again, since cosx=1/2, which is positive...the solutions would be in quadrant I and IV

3...same idea, positive answer cscx = 2/(sqrt3)..solutions are in quadrant I and II

Please help, thanks!

1. The period of the tangent function is $\pi$ , not $2\pi$ .

So you have

$\sqrt{3}\tan{x} + 1 = 0$

$\sqrt{3}\tan{x} = -1$

$\tan{x} = -\frac{1}{\sqrt{3}}$

Now since your solution is in the second and fourth quadrants:

$x = \left\{\pi - \frac{\pi}{6}, 2\pi - \frac{\pi}{6}\right\} + k\pi, k \in \mathbf{Z}$

$x = \left\{ \frac{5\pi}{6}, \frac{11\pi}{6}\right\} + k\pi, k \in \mathbf{Z}$ .

2. $4\cos^2{x} - 1 = 0$

$4\cos^2{x} = 1$

$\cos^2{x} = \frac{1}{4}$

$\cos{x} = \pm \frac{1}{2}$

Since the solution can be in any of the four quadrants:

$x = \left\{\frac{\pi}{3}, \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 2\pi - \frac{\pi}{3}\right\} + 2k\pi, k \in \mathbf{Z}$

$= \left\{\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\right\} + 2k\pi, k \in \mathbf{Z}$ .

3. $3\csc^2{x} - 4 = 0$

$3\csc^2{x} = 4$

$\csc^2{x} = \frac{4}{3}$

$\sin^2{x} = \frac{3}{4}$

$\sin{x} = \pm \frac{\sqrt{3}}{2}$ .

Since the solution can be in any four of the quadrants:

$x = \left\{ \frac{\pi}{3}, \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 2\pi - \frac{\pi}{3}\right\} + 2k\pi, k \in \mathbf{Z}$

$= \left\{\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\right\} + 2k\pi, k \in \mathbf{Z}$ .

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