# Trigonometric Equations

1. Sep 2, 2010

### zorro

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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?

2. Sep 2, 2010

### 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?

3. Sep 2, 2010

### zorro

Thanks....I got it :)