Question

please help!!!Solve sec^2θ + tanθ - 3 = 0 for 0° ≤ θ < 90°.

θ = ???

Answer 1

I think theta = 0 to 90degrees

Answer 2

plot | -3+sec^2(theta)+tan(theta) = 0 | theta = 0 to 90 °

Answer 3

then would it be 45 degrees?? i got that 0.0

Answer 4

\[1+\tan ^2\theta +\tan \theta -3=0\]
\[\tan ^2\theta+\tan \theta -2=0\]
\[(\tan \theta +2)(\tan \theta-1)=0\]
\[\tan \theta=-2 or \tan \theta=1\]
\[\theta=45^{o}\]