Question

the value 5pi/2 is a solution for the equation 2sin^2 x-sin x -1 =0 true or false

Answer 1

this is like saying is pi/2 a solution to the problem,

The curve of y = sin(x) has a period of 2pi.... so once it has travelled 2pi along the x- axis it repeats itself...

so

sin(pi/2) = 1

so you have to substitute 1 for sin(x) and decide if its a solution.

so you have

2*(1)^2 - 1 - 1 = 0

is it a solution or not...?

hope this helps

Answer 2

nope

Answer 3

\[\sin {(2\pi{n}+x)} = \sin{x}\] if n is an integer because 2*pi is one complete revolution around the unit circle. That means we can use pi/2 instead of 5pi/2. sin(pi/2) is 1, because pi/2 is 1 quarter of a revolution, so x = 0, y = 1 and the sin represents the y component (cos is the x component).

So, as @campbell_st put it, we've got 2*(1)^2 - 1 -1 = 0. 2 - 1 - 1 = 0, right? Because that is true, the there is a solution and 5pi/2 fits. As does pi/2, 9pi/2, 13pi/2, etc.

Answer 4

well try this

substitute sin(5pi/2) = 1 into your equation ...