Solve: tan2x = 8cos^2x - cotx

Question

The intervals are: [0, pi/2]

 

Answer 1

tan(2x)=8cos²(x)-cot(x),

2tan(x)/(1-tan²(x))=8cos²(x)-cot(x),

2tan(x)=8cos²(x)-8sin²(x)-cot(x)+tan(x),

tan(x)+cot(x)=8cos(2x),

(sin²(x)+cos²(x))/(sin(x)cos(x))=8cos(2x),

1/(sin(x)cos(x))=8cos(2x),

1=8cos(2x)sin(x)cos(x)=4(2sin(x)cos(x))cos(2x)=4sin(2x)cos(2x)=2sin(4x),

sin(4x)=½, 4x=π/6+2nπ or 4x=5π/6+2nπ, where n is an integer.

x=π/24+nπ/2 or 5π/24+nπ/2 (radians).

x=7.5°+90n° or 37.5°+90n°.

However, if we return to the original question and set x=90°, we get:

tan(2x)=tan(180)=0; cos(90)=0; cot(90)=0 so all the terms are zero, making x=90° another solution.

If x=270°, tan(2x)=tan(540)=0; cos(270)=0; cot(270)=0, so all terms=0, and x=270° is another solution, which can be more generally expressed as x=90(2n+1)°. This is not picked up in the solution method because tan(90(2n+1)) is undefined (infinity), but cot(90(2n+1))=0. The expansion of tan(2x)=2tan(x)/(1-tan²(x)), and when x=90° (or an odd multiple of it), this expression is undefined. The algebraic and arithmetic manipulation that follows is based on tan(x) being defined throughout. Although x=90(2n+1) is a solution, the solution method tacitly bypasses the anomaly in the expansion of tan(2x) and subsequent logic.