Question

If cos^2x = sinx.cosx.355860.y then x = ?

Answer 1

Answer:

sin²(x) + 2sin(x)cos(x) + cos²(x) = y + 1/y

1 + sin(2x) = y + 1/y

If y is real then

y + 1/y ≤ -2

or

y + 1/y ≥ 2 **

1 + sin(2x) ≥ 2

sin(2x) = 1 since sin(2x) cannot be greater than 2

2x = π/2

x = π/4

Check: sin(π/4) + cos(π/4) = √2

√(1 + 1/1) = √2

sin(π/4) + cos(π/4) = √(y + 1/y) for x=π/4, y=1

OR

1 + sin(2x) ≤ -2

sin(2x) ≤ -3

No further solutions

x = π/4

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**

u = y + 1/y

du/dy = 1 - 1/y²

du/dy = 0 at y = {-1,1}

d²u/dy² = 2/y³

d²u/dy² > 0 at y = 1 ---- local minimum

d²u/dy² < 0 at y = -1 ---- local maximum

u(1) = 1 + 1/1 = 2

u(-1) = -1 - 1/1 = -2

Answer 2

Answer:

sin²(x) + 2sin(x)cos(x) + cos²(x) = y + 1/y

1 + sin(2x) = y + 1/y

If y is real then

y + 1/y ≤ -2

or

y + 1/y ≥ 2 **

1 + sin(2x) ≥ 2

sin(2x) = 1 since sin(2x) cannot be greater than 2

2x = π/2

x = π/4

Check: sin(π/4) + cos(π/4) = √2

√(1 + 1/1) = √2

sin(π/4) + cos(π/4) = √(y + 1/y) for x=π/4, y=1

OR

1 + sin(2x) ≤ -2

sin(2x) ≤ -3

No further solutions

x = π/4

--------------------------------------------------------------------------------------

**

u = y + 1/y

du/dy = 1 - 1/y²

du/dy = 0 at y = {-1,1}

d²u/dy² = 2/y³

d²u/dy² > 0 at y = 1 ---- local minimum

d²u/dy² < 0 at y = -1 ---- local maximum

u(1) = 1 + 1/1 = 2

u(-1) = -1 - 1/1 = -2