Question

tan^(2)θ + 3secθ+3=0

Answer 1

\[1+\tan^2 \theta=\sec^2 \theta\]
So
\[\tan^2\theta=\sec^2 \theta -1\]
Put this in the given equation
\[\sec^2\theta-1+3\sec \theta+3=0\]
\[\sec^2\theta+3\sec \theta+2=0\]
Put \[\sec \theta=x\]
Now you have\[x^2+3x+2=0\]
Solve for x, can you do that?

Answer 2

Yep I just got it

Answer 3

Finished it, just as you posted that. Tyvm tho :)

Answer 4

My Final Answers were: cos = -1/2, cos = -1

Answer 5

No problem :D

Answer 6

What a clear explanation @ash2326

Answer 7

Which converts to 120, 180, and 240 degrees. Got it =D

Answer 8

Remember the identity:

\[\sec^2\theta=\tan^2\theta+1\rightarrow \tan^2\theta=\sec^2\theta-1\]

We re-write the equation as:

\[(\sec^2(\theta) -1)+3\sec(\theta)+3=0\]\[\sec^2(\theta)+3\sec(\theta)+2=0\]This is a quadratic and so just factor it to get the possible values of sec(x).

\[(\sec(\theta)+1)(\sec(\theta)+2)=0\]

From the factorisation, either sec(x) + 1 = 0 or sec(x) + 2 = 0. Now we just solve for the two cases and get our values.

Case 1:
\[\sec(\theta)=-1\rightarrow \cos(\theta)=-1\]\[\theta=\pi\]Case 2:\[\sec(\theta)=-2\rightarrow \cos(\theta)=-\frac{1}{2}\]\[\theta=\frac{2\pi }{ 3 },\frac{ 4\pi }{ 3}\]\[\therefore \theta=\pi,\frac{ 2\pi }{ 3 },\frac{ 4\pi }{ 3 }\]