C3 Trig

I came to get tan^2 0 = 49/576. So to get tan^0 as the question is asking, I square root it and end up with + or - 7/24. My question is how do I know which one is the right one? The answer in the textbook is tan 0 = - 7/24.Theta is meant to be reflex (180-360) but how do I apply this?

I am not sure why you have $\tan ^2\theta$ but

You know that the angle is between 180 and 360

But you also know that $\cos \theta >0$ so this tells you if you are in 180-270 OR 270 -360

I got tan^2 0 by turning cos 0 into sec 0 then sec^2 0 which is identical to (1 + tan^0) then I eventually had tan 0 on its own. It gives the right answer; but looking at the answer this was all stupid because they simply drew up a right angled triangle (with the known 24 and 25 values from cos theta) and simply worked out the tan value from there. But I understand now, cos 0 > 0 means it would have to be in the 4th quadrant and the solution for tan 0 would be negative. Thank you very much for your help

I came to get tan^2 0 = 49/576. So to get tan^0 as the question is asking, I square root it and end up with + or - 7/24. My question is how do I know which one is the right one? The answer in the textbook is tan 0 = - 7/24.Theta is meant to be reflex (180-360) but how do I apply this?

Would someone be able to help me with some C3 Trig?

I have 2 questions

a) √3 sin2(x) + 2sin^2(x) = 1

b) 4 tan2(x) tan(x) =1

Thank you.