Use the maximum, minimum and zero values of sine and cosine functions to solve the equation.

You notice that if cos x=0, the factor `(v*cos x)/(g) = 0` . This 0 factor cancels out the entire product. The cosine is zero if sin x = `+-1` (`x=pi/2 + n*pi/2` ).

If cos x...

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Use the maximum, minimum and zero values of sine and cosine functions to solve the equation.

You notice that if cos x=0, the factor `(v*cos x)/(g) = 0` . This 0 factor cancels out the entire product. The cosine is zero if sin x = `+-1` (`x=pi/2 + n*pi/2` ).

If cos x = -1 => sin x = 0. Put these values in your equation:

`y = -(v/g)*[v*0 +sqrt( (v^2*0^2) + (2*g*h) )]`

`y = -(v*sqrt(2gh))/g`

If `cos x = 1 =gt sin x = 0 =gt y = -(v*sqrt(2gh))/g`

**ANSWER: If sin x is maximal or minimal, the equation cancels out, therefore the roots of the equation are x=`pi/2 + n*pi/2` , n=1,2,...**