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~csc^2(x) = 3~cot^2(x) - 1[/tex]
$\displaystyle \frac{2}{sin^2(x)} = \frac{3~cos^2(x)}{sin^2(x)} - 1$
$\displaystyle 2 = 3~cos^2(x) - sin^2(x)$
and since $\displaystyle cos^2(x) = 1 - sin^2(x)$
$\displaystyle 2 = 3 - 3~sin^2(x) - sin^2(x)$
$\displaystyle -1 = -2~sin^2(x)$
You can finish from here.
-Dan
For the first problem, write it as $\displaystyle \frac{2}{\sin ^2x} = \frac{3 \cos ^2x}{\sin ^2x} - \frac{\sin ^2x}{\sin ^2x}$. Then, as long as sin x isn't 0, you have $\displaystyle 2 = 3 \cos ^2x - \sin ^2x$, which you can then arrange as:
$\displaystyle 2 = 3 \cos ^2x - (1 - \cos ^2x)$
$\displaystyle 2 = 3 \cos ^2x - 1 + \cos ^2x$
$\displaystyle 3 = 4 \cos ^2x$
$\displaystyle \cos ^2x = \frac{3}{4}$
New questions go in new threads.
$\displaystyle 2~cos^2(x) - cos(x) = 2 - \frac{1}{cos(x)}$
$\displaystyle 2~cos^3(x) - cos^2(x) - 2~cos(x) + 1 = 0$
Again, letting y = cos(x) and solving the resulting polynomial I get that
$\displaystyle cos(x) = 1, cos(x) = -1, cos(x) = 1/2$
-Dan
He means substituting y = cos(x)
$\displaystyle 2 cos^3(x)-cos^2(x)-2cos(x)+1 = 0 \Rightarrow
2y^3-y^2-2y+1 = 0$
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
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