Is There a Solution for Trigonometric Equations with 1+cosx and sin^2x?

[zorro]

Homework Statement

Solve (1+cosx)sin^2x = x^2 + 1/(x^2)

Homework Equations

The Attempt at a Solution

RHS is always greater than or equal to 2
LHS is always less than or equal to 2 as
1+cosx is always less than or equal to 2 and sin^2x is always less than or equal to 1

so there is a solution when LHS=RHS=2
but my book says there is no solution as LHS is always less than 2 (not equal to 2)

How?

 

[JonF]

You’re off to a good start.

Let y = cosx and z = sin^2(x), both of these are inclusively bound between 0 and 1, Your LFH then becomes (1+y)z, where 0 <= y <= 1; 0 <= z <= 1. So what value does y and z need to be for the LHS to equal 2?

So this puts your equation under the constraints of cosx = 1; sin^2(x) = 1; x^2 + x^-2 = 2

Note sin^2(x) = 1 => sin(x) = +/-1

When does cos(x) = 1? When does sin(x) = =/-1

Can you ever meet both of these constraints with the same value of x?

 

[zorro]

Thanks...I got it :)