2sec^2(x) = 3 − tan(x)

Question

Solve for the exact solutions in the interval $[0,2\pi )$

Answer 1

Rewrite this as:

2(1+tan²(x))=3-tan(x),

2+2tan²(x)=3-tan(x),

2tan²(x)+tan(x)-1=0=(2tan(x)-1)(tan(x)+1).

The clue in this type of question is to remember which quadrant contains the negative and positive values for the trig functions. Quadrant 1: 0-π/2, all positive; quadrant 2: π/2-π, sine positive only; quadrant 3: π-3π/2, tangent positive only; quadrant 4: 3π/2-2π, cosine positive only.

So tan(x)=½ or -1, x=tan^-1(0.5)=0.4636 or π+0.4636=3.6052; or tan^-1(-1)=7π/4, 3π/4.

Solution: x=0.4636, 3π/4, 3.6052, 7π/4.