1. ## Solve for theta

Write all positive values within the interval of 0 degrees < $\theta$ < 270 degrees

$1 - sec^{2} 2\theta - 3tan 2\theta = 0$

2. Originally Posted by reiward
Write all positive values within the interval of 0 degrees < $\theta$ < 270 degrees

$1 - sec^{2} 2\theta - 3tan 2\theta = 0$
Substitute $\sec^2 (2 \theta) = \tan^2 (2 \theta) + 1$, simplify and solve the resulting simple trig equation.

3. Originally Posted by mr fantastic
Substitute $\sec^2 (2 \theta) = \tan^2 (2 \theta) + 1$, simplify and solve the resulting simple trig equation.
$
1 - sec^{2} 2\theta - 3tan 2\theta = 0$

$
1 - (\tan^2 (2 \theta) + 1) - 3tan 2\theta = 0$

$\tan^2 (2 \theta) - 3tan 2\theta = 0$

Im stuck. What do I factor it to?

4. Originally Posted by reiward
$
1 - sec^{2} 2\theta - 3tan 2\theta = 0$

$
1 - (\tan^2 (2 \theta) + 1) - 3tan 2\theta = 0$

$\tan^2 (2 \theta) - 3tan 2\theta = 0$ Mr F says: This is wrong. There is a careless mistake in signs.

Im stuck. What do I factor it to?
Factorise. Note that $\tan (2 \theta)$ is a common factor. Then use the Null Factor Law to get two simple trig equations. Solve them.

5. Originally Posted by mr fantastic
Substitute $\sec^2 (2 \theta) = \tan^2 (2 \theta) + 1$, simplify and solve the resulting simple trig equation.
Originally Posted by mr fantastic
Factorise. Note that $\tan (2 \theta)$ is a common factor. Then use the Null Factor Law to get two simple trig equations. Solve them.
Oh okay.

$-\tan^2 (2 \theta) - 3tan 2\theta = 0$

Can I make that to $\tan^2 (2 \theta) + 3tan 2\theta = 0$ ?

If so, would the factor be $\tan(2 \theta) [tan + 3]= 0$?

6. Originally Posted by reiward
Oh okay.

$-\tan^2 (2 \theta) - 3tan 2\theta = 0$

Can I make that to $\tan^2 (2 \theta) + 3tan 2\theta = 0$ ?

If so, would the factor be $\tan(2 \theta) [tan + 3]= 0$?
Yes.
Yes.