Question

If sinx + cosx = (y + 1/y)^1/2, y >0 and x belongs to [0,π]then find the least positive value of x satisfying the given condition ?​

Answer 1

Answer:

$\frac{\pi}{4}$

Step-by-step explanation:

y > 0

Use AM - GM inequality

${y + \frac{1}{y} \over2} \geqslant \sqrt{y \times \frac{1}{y} } \\ y + \frac{1}{y} \geqslant 2$

Now it results

$\sin(x) + \cos(x) \geqslant \sqrt{2}$

For minimum value of x .

$\frac{1}{ \sqrt{2} } \sin(x) + \frac{1}{ \sqrt{2} } \cos(x) \geqslant 1 \\ \sin( \frac{\pi}{4} + x) \geqslant 1$

It is only possible if

$\sin( \frac{\pi}{4} + x) = 1 \\ \frac{\pi}{4} + x = (4n + 1) \frac{\pi}{2} \\ x = 2n\pi + \frac{\pi}{4}$

So possible value in [ 0 , π ] is π/4 .