Solve: tan2x = 8cos^2x - cotx

Question

with the intervals: [ 0 ≤ x ≤ pi/2 ]

Answer 1

tan(2x)=2tan(x)/(1-tan²(x)), provided tan(x)≠±1, that is, x≠±π/4; note that when x=π/2, tan(x) is not defined but tan(2x)=0, cot(x)=0 and cos(x)=0.

2tan(x)/(1-tan²(x))=8cos²(x)-cot(x),

2tan(x)=(8cos²(x)-cot(x))(1-tan²(x)),

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

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

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

x=π/24+nπ/2, 5π/24+nπ/2. When n=0, x=π/24, 5π/24. When n=1, x=13π/24 which is outside the given interval.

However, returning to the original question we excluded x=±π/4, that is, 2x=±π/2. And it was shown that tan(2x)=cot(x)=cos(x)=0 when x=π/2. The complete solution is:

 x=π/24, 5π/24, π/2 in the given interval.