sin^2x=5 has how many solutions on the interval 0 less than or equal to x less than or equal to 2pi?

Question

A.0

B.2

C.1

D.Infinity

 

I need a step by step on how to do this.

Answer 1

Our interval is    0 ≤ x ≤ 2π.

 

 

sin²(x) = 5

 

 

Square-root both sides of equation.

 

sin(x) = ±√(5)

 

sin(x) = √(5)           and            sin(x) = -√(5)

 

sin(x) = 2.2             and            sin(x) = -2.2

 

x = sin^-1(2.2)          and               x = sin^-1(-2.2)

 

The highest value of sine we can have is 1.  The lowest value of sine we can have is -1.  Since our sine is greater than 1 and the other value of sine is less than -1, there are no solutions.