Please Help!!! I'm not sure how to solve this.

Question

how would i solve this equation for x where pi ≤ x ≤ 2pi?

sin^2x-cos^2x ÷ cos^2x=0.

it says that the answers should be entered as fractions.

i know the steps but i can't figure out where i am going wrong so please, if you help, be clear and concise.

thank you in advance.

Answer 1

If sin² x = 1 then

sin x = ±1 then

x = π/2 or x = 3π/2

π ≤ x ≤ 2π then

x = 3π/2

Answer 2

pi ≤ x ≤ 2pi

sin^2x-cos^2x ÷ cos^2x=0.

sin^2x-1=0

sin^2x=1

sin90°=1^2x=1

x=any amount

90°>2pi. So x is between 3.14 and 6.24.