# Thread: Finding exact solutions in the interval

1. ## Finding exact solutions in the interval

Hello,

I am having trouble with this question, it say's:

Find all exact solutions in the interval [0,2[pi]).

a.) tan x= -2sin x

b.) sin^2 θ + 5cosθ +1=0

c.) sin^2 θ = -cos2θ

Any help would be greatly appreciated =)

2. Originally Posted by KAY03
Hello,

I am having trouble with this question, it say's:

Find all exact solutions in the interval [0,2[pi]).

a.) tan x= -2sin x

b.) sin^2 θ + 5cosθ +1=0

c.) sin^2 θ = -cos2θ

Any help would be greatly appreciated =)
a) $\tan{x} = -2\sin{x}$

$\frac{\sin{x}}{\cos{x}} = -2\sin{x}$

$\frac{\sin{x}}{\sqrt{1 - \sin^2{x}}} = -2\sin{x}$

$\frac{\sin^2{x}}{1 - \sin^2{x}} = 4\sin^2{x}$

$\sin^2{x} = 4\sin^2{x}(1 - \sin^2{x})$

$\sin^2{x} = 4\sin^2{x} - 4\sin^4{x}$

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

$0 = \sin^2{x}(3 - 4\sin^2{x})$

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

Case 1:

$\sin^2{x} = 0$

$\sin{x} = 0$

$x = \left\{0, \pi\right\} + 2\pi n$ where $n \in \mathbf{Z}$.

Case 2:

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

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

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

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

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

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

Putting it all together, we find

$x = \left\{0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3} \right\} + 2\pi n$ where $n \in \mathbf{Z}$.

In the interval $x \in [0, 2\pi)$, the solutions are:

$x = \left\{0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3} \right\} + 2\pi n$ where $n \in \mathbf{Z}$.

You may want to check these answers against the graph of $\tan{x} + 2\sin{x}$ though. Since I obtained the answers through squaring, it's possible that there are extraneous roots.

3. Originally Posted by KAY03
Hello,

I am having trouble with this question, it say's:

Find all exact solutions in the interval [0,2[pi]).

a.) tan x= -2sin x

b.) sin^2 θ + 5cosθ +1=0

c.) sin^2 θ = -cos2θ

Any help would be greatly appreciated =)
(b) $\sin^2{\theta} + 5\cos{\theta} + 1 = 0$

$1 - \cos^2{\theta} + 5\cos{\theta} + 1 = 0$

$-\cos^2{\theta} + 5\cos{\theta} + 2 = 0$

$\cos^2{\theta} - 5\cos{\theta} - 2 = 0$.

If we let $X = \cos^{\theta}$ we have a quadratic:

$X^2 - 5X - 2 = 0$

$X^2 - 5X + \left(-\frac{5}{2}\right)^2 - \left(-\frac{5}{2}\right)^2 - 2 = 0$

$\left(X - \frac{5}{2}\right)^2 - \frac{25}{4} - \frac{8}{4} = 0$

$\left(X - \frac{5}{2}\right)^2 - \frac{33}{4} = 0$

$\left(X - \frac{5}{2}\right)^2 = \frac{33}{4}$

$X - \frac{5}{2} = \pm \frac{\sqrt{33}}{2}$

$X = \frac{5 \pm \sqrt{33}}{2}$.

So $\cos{\theta} = \frac{5 + \sqrt{33}}{2}$ or $\cos{\theta} = \frac{5 - \sqrt{33}}{2}$.

You will need to use technology to find $\theta$ now.

4. Originally Posted by KAY03
Hello,

I am having trouble with this question, it say's:

Find all exact solutions in the interval [0,2[pi]).

a.) tan x= -2sin x

b.) sin^2 θ + 5cosθ +1=0

c.) sin^2 θ = -cos2θ

Any help would be greatly appreciated =)
(c) $\sin^2{\theta} = -\cos{2\theta}$

$\sin^2{\theta} = -(\cos^2{\theta} - \sin^2{\theta})$

$\sin^2{\theta} = \sin^2{\theta} - \cos^2{\theta}$

$0 = -\cos^2{\theta}$

$0 = \cos^2{\theta}$

$0 = \cos{\theta}$

$\theta = \frac{\pi}{2} + \pi n$ where $n \in \mathbf{Z}$.

So in the interval $\theta \in [0, 2\pi)$

$\theta = \left\{\frac{\pi}{2}, \frac{3\pi}{2} \right\}$.