Question

Solve the equation:

Answer 1

\[5 \sec^22\theta=\tan2\theta + 9 \]
Give all solutions between 0-180 degrees.


Yet another one I am stuck at, any help appreciated.

Answer 2

\(1+\tan^2\theta=\sec^2\theta\)

Answer 3

\(5(1+\tan^22\theta)=\tan2\theta+9\)

Answer 4

Thank you, will have a go and see if I can get it done now. Was using sec=1/cos earlier.

Answer 5

Got it, thanks again @ajprincess.
Hopefully I can pass my finals next week :)

Answer 6

u r welcome. best of luck for ur finals.:)

Answer 7

\[\cos^2(\theta) + \sin^2(\theta) = 1\]
Divide by \[\cos^2(2\theta)\] we get
\[1 + \tan^2(2\theta) = \sec^2(2\theta)\]
With this result, plug it in your equation:
\[5\left( 1+ \tan^2(2\theta) \right) = \tan(2\theta) + 9\]
then
\[5\tan^2(2\theta) - \tan(2\theta) - 4 = 0\]
That's a quadratic equation on \[\tan(2\theta)\], just solve it:
\[\tan(2\theta) = -4/5\] or \[\tan(2\theta) = 1\]
for the first one:
\[\theta = \frac{ \tan^{-1} (-4/5) }{ 2 } = -19.33º\]
for the second one:
\[\theta = \frac{ \tan^{-1} (1) }{ 2 } = (1/8)(4\pi n + \pi)\]
Set n = 0 and n = 1:
\[\theta = 22.5º\] and \[\theta = 112.5º\]