Thread: Need help on a couple of Trig questions, assignment due tomorrow

1. Need help on a couple of Trig questions, assignment due tomorrow

I have an assignment thats due tomorrow, heres a couple I'm not sure how to do.

1)Factor & simplify: Tan^(3)x+8

2) Sec^(2)x-2Tan^(2)x=0
I got x= pi/4, 3pi/4, and 5pi/4 but I don't know if it's right

3) Tanx=-sqrt(3)/3
I got x=-pi/6 but not sure on answer

2. Originally Posted by KURO
I have an assignment thats due tomorrow, heres a couple I'm not sure how to do.

1)Factor & simplify: Tan^(3)x+8

2) Sec^(2)x-2Tan^(2)x=0
I got x= pi/4, 3pi/4, and 5pi/4 but I don't know if it's right

3) Tanx=-sqrt(3)/3
I got x=-pi/6 but not sure on answer

1)Factor & simplify: Tan^(3)x+8
That could be a typo.
I see that it should be tan^(3)x = 8
Then, take the cube roots of both sides,
tanX = 2 ......a positive tan value, so X must be in the 1st or 3rd quadrants.
X = arctan(2)
So,
X = 1.10715 and 4.24874 radians

-----------------------------------
2) Sec^(2)x-2Tan^(2)x = 0

In the trig identity
sin^2(X) +cos^2(X) = 1,
Divide both sides by cos^2(X),
tan^2(X) +1 = sec^2(X) -------**

So,
tan^2(X) +1 -2tan^2(X) = 0
-tan^2(X) = -1
tan^2(X) = 1
tanX = +,-1

When tanX = 1,
positive tan value, so X is in the 1st and 3rd quadrants,
X = arctan(1)
X = pi/4 and (pi +pi/4) = 5pi/4
or, X = pi/4, and 5pi/4

When tanX = -1,
negative tan value, so X is in the 2nd and 4th quadrants,
X = arctan(-1)
X = (pi -pi/4) = 3pi/4 in the 2nd quadrant
X = (2pi -pi/4) = 7pi/4 in the 4th quadrant

Therefore, X = pi/4, 3pi/4, 5pi/4 and 7pi/4

----------------------------------
3) Tanx = -sqrt(3)/3

If you divide both numerator and denominator by sqrt(3), that's the same as
tanX = -1 / sqrt(3)
Negative tan, so X is in the 2nd and 4th quadrants.
X = arctan[-1 / sqrt(3)]
X = (pi -pi/6) = 5pi/6 in the 2nd quadrant
X = (2pi -pi/6) = 11pi/6 in the 4th quadrant
So,
X = 5pi/6 and 11pi/6

3. Originally Posted by ticbol
1)Factor & simplify: Tan^(3)x+8
That could be a typo.
I see that it should be tan^(3)x = 8
Then, take the cube roots of both sides,
tanX = 2 ......a positive tan value, so X must be in the 1st or 3rd quadrants.
X = arctan(2)
So,
X = 1.10715 and 4.24874 radians

-----------------------------------
2) Sec^(2)x-2Tan^(2)x = 0

In the trig identity
sin^2(X) +cos^2(X) = 1,
Divide both sides by cos^2(X),
tan^2(X) +1 = sec^2(X) -------**

So,
tan^2(X) +1 -2tan^2(X) = 0
-tan^2(X) = -1
tan^2(X) = 1
tanX = +,-1

When tanX = 1,
positive tan value, so X is in the 1st and 3rd quadrants,
X = arctan(1)
X = pi/4 and (pi +pi/4) = 5pi/4
or, X = pi/4, and 5pi/4

When tanX = -1,
negative tan value, so X is in the 2nd and 4th quadrants,
X = arctan(-1)
X = (pi -pi/4) = 3pi/4 in the 2nd quadrant
X = (2pi -pi/4) = 7pi/4 in the 4th quadrant

Therefore, X = pi/4, 3pi/4, 5pi/4 and 7pi/4

----------------------------------
3) Tanx = -sqrt(3)/3

If you divide both numerator and denominator by sqrt(3), that's the same as
tanX = -1 / sqrt(3)
Negative tan, so X is in the 2nd and 4th quadrants.
X = arctan[-1 / sqrt(3)]
X = (pi -pi/6) = 5pi/6 in the 2nd quadrant
X = (2pi -pi/6) = 11pi/6 in the 4th quadrant
So,
X = 5pi/6 and 11pi/6
Thanks for the help! But for Number 1, that isn't a typo its: Tan^(3)x+8. Any idea how to solve it?

4. Originally Posted by KURO
Thanks for the help! But for Number 1, that isn't a typo its: Tan^(3)x+8. Any idea how to solve it?
If it were not a typo, then,
tan^3(X) +8
It is a sum of two cubes.
= tan^3(X) +2^3
= [tanX +2]*[tan^2(X) -2tanX +2^2]
= (tanX +2)(tan^2(X) -2tanX +4)

That's it. It's just factored.

5. Thanks!

6. Originally Posted by kuro
i have an assignment thats due tomorrow, heres a couple i'm not sure how to do.

1)factor & simplify: Tan^(3)x+8

hint ... A^3 + b^3 = (a + b)(a^2 - ab + b^2)

2) sec^(2)x-2tan^(2)x=0
i got x= pi/4, 3pi/4, and 5pi/4 but i don't know if it's right

missing x = 7pi/4

3) tanx=-sqrt(3)/3
i got x=-pi/6 but not sure on answer

what interval for solutions? 0 to 2pi? If so, missing x = 5pi/6 and 11pi/6