Question

5tanx-2cotx+1=0...... .

Answer 1

Step-by-step explanation:

5(tanx)^2 -2 +tanx=0

let tanx=y

5y^2 + y-2=0

y=0.5403 or y= -0.7403

tanx=0.5403

x=arctan(0.5403)

x=28.38°

or

tanx=-0.7403

x=arctan(-0.7403)

x=143.49°