Get the solution to "Do the equation2cos2(x2)sin2(x2)=x2+1x2"

Emilia Hoffman

Answered question

2022-04-14

Do the equation

2 \cos^{2} (\frac{x}{2}) \sin^{2} (\frac{x}{2}) = x^{2} + \frac{1}{x^{2}}

tempur8x43

Beginner2022-04-15Added 16 answers

the right hand of the equation, you have

x^{2} + \frac{1}{x^{2}} = {(x - \frac{1}{x})}^{2} + 2 \geq 2

(1)
and equality occurs for

x = \pm 1

on the left hand side, we have

\frac{1}{2} \sin^{2} x = 2 \cos^{2} (\frac{x}{2}) \sin^{2} (\frac{x}{2}) \leq \frac{1}{2}

(2)
Equality for

\sin x = \pm 1

but (1) and (2) are inconsistent, therefore

2 \cos^{2} (\frac{x}{2}) \sin^{2} (\frac{x}{2}) = x^{2} + \frac{1}{x^{2}}

has no real solutions.

Colin Collins

Beginner2022-04-16Added 10 answers

In the same spirit as baharampuri's answer, the equation write

\frac{\sin^{2} (x)}{2} = x^{2} + \frac{1}{x^{2}}

The lhs is almays

< \frac{1}{2}

while the rhs is

\geq 2

For the last point, consider

y = x^{2} + \frac{1}{x^{2}}

y^{'} = 2 x - \frac{2}{x^{3}}

y^{″} = 2 + \frac{6}{x^{4}}

So,

y^{'} = 0

for

x = \pm 1

and at this point

y = 2

. The second derivative test proves that this is a minimum.

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