# Thread: Help (Trig Equations) (ugh)

1. ## Help (Trig Equations) (ugh)

1. 3cot^2 (x) - 1 = 0
My answer: pi/3, 2pi/3, 4pi/3, 5pi/3

2. 4cos^2 (x) - 1 = 0
My answer: pi/3, 2pi/3, 5pi/3, 4pi/3

3. 2sin (x) + csc (x) = 0
i got to the part: sin^2 (x) = -1/2

4. 4sin^3 (x) + 2sin^2 (x) - 2sin (x) = 1
(ahhhhhh)

2. You're kidding on #4, right? Don't those middle terms cancel each other and leave a much simpler problem?

3. EDIT: Nvm, sorry.

4. woops my bad i fixed it lol

5. Hello, NeedHelp18!

Ya done good!

$\displaystyle 1)\;\;3\cot^2\!x - 1 \:=\: 0$
My answers: .$\displaystyle \frac{\pi}{3},\;\frac{2\pi}{3},\;\frac{4\pi}{3},\; \frac{5\pi}{3}$ . . . . Right!

$\displaystyle 2)\;\;4\cos^2\!x - 1 \:=\: 0$
My answer: .$\displaystyle \frac{\pi}{3},\;\frac{2\pi}{3},\;\frac{5\pi}{3},\; \frac{4\pi}{3}$ . . . . Yes!

$\displaystyle 3)\;\;2\sin x + \csc x \:= \:0$
My answer: unknown ... LOL . . . . Correct!
i got to the part: .$\displaystyle \sin^2\!x \:=\:-\frac{1}{2}$ . . . . There are no real roots

$\displaystyle 4)\;\;4\sin^3\!x + 2\sin^2\!x - 2\sin x \:= \:1$
We have: .$\displaystyle 4\sin^3\!x + 2\sin^2\!x - 2\sin x - 1 \;=\;0$

Factor: .$\displaystyle 2\sin^2\!x(2\sin x + 1) - (2\sin x + 1) \;=\;0$

Factor: .$\displaystyle (2\sin x + 1)(2\sin^2\!x - 1) \;=\;0$

. . $\displaystyle \begin{array}{ccccccc}2\sin x +1\:=\:0 & \Rightarrow & \sin x \:=\:-\frac{1}{2} & \Rightarrow & x \:=\:\dfrac{7\pi}{6},\:\dfrac{11\pi}{6} \\ \\[-3mm] 2\sin^2\!x-1\:=\:0 & \Rightarrow & \sin x \:=\:\pm\frac{1}{\sqrt{2}} & \Rightarrow & x \:=\:\dfrac{\pi}{4},\:\dfrac{3\pi}{4},\:\dfrac{5\p i}{4},\:\dfrac{7\pi}{4} \end{array}$

6. Sometime i amaze myself
i was so sure those answers were wrong

7. . . . . There are no real roots

Is it because it would be an imaginary number?

8. Originally Posted by NeedHelp18
. . . . There are no real roots

Is it because it would be an imaginary number?
Yes it is =)