cos^2x-5sin(x)+5=0?

　
cos²x − 5sinx + 5 = 0
(1−sin²x) − 5sinx + 5 = 0
−sin²x − 5sinx + 6 = 0
sin²x + 5sinx − 6 = 0
(sinx + 6) (sinx − 1) = 0

sinx = −6 -----> no solution

sinx = 1 ----> x = π/2

Cos^2x-5sin(x)+5=0?
Use inverse functions where needed to find all solutions (if they exist) of the given equation on the interval [0,2pi).
Thanks!!!

Nope, no inverse functions are required to do this problem.


cos^2(x) - 5 sin (x) + 5 = 0

Use sin^2(x) + cos^2(x) = 1 hence cos^2(x) = 1 - sin^2(x)

(1 - sin^2(x)) - 5 sin (x) + 5 = 0
1 - sin^2(x) - 5 sin (x) + 5 = 0

Combine like terms.

-sin^2(x) - 5 sin (x) + 6 = 0

Factor out an -1 out of the terms..,.

- (sin^2(x) + 5 sin (x) - 6) = 0

Now look for two factors that multiply to the last term (-6) and adds to the middle term (5) and then set them into the form (sin x + a)(sin x + b)..
Those two factors are -1 and 6...so factoring yields..

- (sin x - 1)(sin x + 6) = 0

Now set each factor equal to 0 and solve for x...

sin x - 1 = 0
sin x - 1 + 1 = 0 + 1
sin x = 1

Now looking at the unit circle in the domain [0, 2pi) where y-coordinate is 1...that angle is π/2...

sin x + 6 =0
sin x + 6 - 6 = 0 - 6
sin x = -6
No solution...range of sin x is -1 <= x <= 1 (or domain of arcsin is -1 <= x <= 1)..


So only solution in this interval is x = π/2