Question

interval [0,2pi)
sinx^2 - 6cosx+6=0

Answer 1

solve

Answer 2

in the future you should write "sin squared" as sin^2(x) or (sinx)^2

Answer 3

(sin^2x)=1-cos^2x
1-cos^2x-6cosx+6=0
-cos^2x-6cosx+7=0
cos^2x+6cosx-7=0
let u=cosx
u^2+6u-7=0
(u-1)(u+7)=0
cosx=1
cosx=-7

Answer 4

so solve both of those equtions

Answer 5

if cosx=1 then x=0
cosx can't be -7
so x=0 is only solution

Answer 6

in the interval [0,2pi)

Answer 7

thankyou

Answer 8

if you like my answer give me a medal :)