# Thread: Help with Trig Equation!

1. ## Trig Equation! Help

Does anyone know how to do this? I have no idea how to even start this. Thanks a lot

2csc^2 x = 3cot^2 x - 1
sin^2x + sin x - 1 = 0
sin x + cos x = 0

2. Originally Posted by NeedHelp18
Does anyone know how to do this? I have no idea how to even start this. Thanks a lot

2csc^2 x = 3cot^2 x - 1
2~csc^2(x) = 3~cot^2(x) - 1[/tex]

$\frac{2}{sin^2(x)} = \frac{3~cos^2(x)}{sin^2(x)} - 1$

$2 = 3~cos^2(x) - sin^2(x)$

and since $cos^2(x) = 1 - sin^2(x)$

$2 = 3 - 3~sin^2(x) - sin^2(x)$

$-1 = -2~sin^2(x)$

You can finish from here.

-Dan

3. Originally Posted by NeedHelp18
Does anyone know how to do this? I have no idea how to even start this. Thanks a lot

2csc^2 x = 3cot^2 x - 1
sin^2x + sin x - 1 = 0
sin x + cos x = 0
For the first problem, write it as $\frac{2}{\sin ^2x} = \frac{3 \cos ^2x}{\sin ^2x} - \frac{\sin ^2x}{\sin ^2x}$. Then, as long as sin x isn't 0, you have $2 = 3 \cos ^2x - \sin ^2x$, which you can then arrange as:

$2 = 3 \cos ^2x - (1 - \cos ^2x)$

$2 = 3 \cos ^2x - 1 + \cos ^2x$

$3 = 4 \cos ^2x$

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

4. Originally Posted by NeedHelp18
sin^2x + sin x - 1 = 0
Let $y = sin(x)$. This transforms the equation to
$y^2 + y - 1 = 0$

Solve for y and then set that equal to sin(x).

-Dan

5. Originally Posted by NeedHelp18
sin x + cos x = 0
If you honestly can't get this one you are thinking too hard about it.
$sin(x) + cos(x) = 0$

$sin(x) = -cos(x)$

$\frac{sin(x)}{cos(x)} = -1$

$tan(x) = -1$
and go from there.

-Dan

6. Originally Posted by NeedHelp18
Does anyone know how to do this? I have no idea how to even start this. Thanks a lot

2csc^2 x = 3cot^2 x - 1
sin^2x + sin x - 1 = 0
sin x + cos x = 0
For the second problem, substitute $\sin x = z$ and then solve the equation $z^2 + z - 1 = 0$ using the quadratic formula. Once you get the two values for z, see if either of them make sense for $\sin x$.

7. Thanks! I was definitely trying too hard lol
My teacher refused to show our class how to do them because she got mad at us for talking soo much, even though i wasnt talking at all lol

8. okay so i tried to do this one and i got half right.
i got pi/3 and 5pi/3.
why does the book also say that i was supposed to get 0 and pi??

here it is:

2cos^2 x - cos x = 2 - sec x

9. Originally Posted by NeedHelp18
okay so i tried to do this one and i got half right.
i got pi/3 and 5pi/3.
why does the book also say that i was supposed to get 0 and pi??

here it is:

2cos^2 x - cos x = 2 - sec x
New questions go in new threads.

$2~cos^2(x) - cos(x) = 2 - \frac{1}{cos(x)}$

$2~cos^3(x) - cos^2(x) - 2~cos(x) + 1 = 0$

Again, letting y = cos(x) and solving the resulting polynomial I get that
$cos(x) = 1, cos(x) = -1, cos(x) = 1/2$

-Dan

10. Originally Posted by topsquark
New questions go in new threads.

$2~cos^2(x) - cos(x) = 2 - \frac{1}{cos(x)}$

$2~cos^3(x) - cos^2(x) - 2~cos(x) + 1 = 0$

Again, letting y = cos(x) and solving the resulting polynomial I get that
$cos(x) = 1, cos(x) = -1, cos(x) = 1/2$

-Dan
how do you do this part?
im trying it but nothing.

11. He means substituting y = cos(x)

$2 cos^3(x)-cos^2(x)-2cos(x)+1 = 0 \Rightarrow
2y^3-y^2-2y+1 = 0$

and then solving that equation.

To solve a polynomial, find a solution to that equation (in this case, y = 1 satisfies the equation) Therefore its a solution to the polynomial and (y - 1) is a factor. Divide the polynomial by that factor, and you should get a quadratic. Apply the quadratic formula to that to get the other two answers.

Remember, when you have your solutions, don't forget to change your y back into cos(x)

For a more indepth look at solving polynomails look at this site Solving Polynomial Equations

12. so it becomes
2y^3 - y^2 - 2y + 1 = 0

solve for y as you would a polynomial then sub back into the y

i get

2 cos theta -1 = 0

and

cos theta + 1 = 0

and

cos theta -1 = 0

depends what range the solution is required

between 0 and 2 pi (inclusive)

i get: 0, pi/3, pi, 5pi/3, 2pi