Can you find this tangents equation?

Calculus Level 4

Find the straight line equation that is tangent of the curve cos ⁡ ( x 3 + 2 y ) = ( 2017 − ( ln ⁡ 2 ) ⋅ π ) x + π ⋅ 2 x \cos(x^3 + 2y) = (2017 - (\ln2 )\cdot\pi ) x + \pi \cdot2^x at the point ( x , y ) = ( 0 , π 2 ) (x,y) = \left( 0 ,\frac\pi2 \right) (if any).

there is no tangents equation y = ∞ y=\infty y = 0 y=0 x = ∞ x=\infty x = 0 x=0

Uzumaki Nagato Tenshou Uzumaki by uzumaki nagato tenshou u… , 26

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1 solution

Tom Engelsman
Feb 7, 2017

If one looks at this curve at x = 0 , x = 0, the result is c o s ( 2 y ) = π cos(2y) = \pi . This is impossible for all real y y since c o s ( 2 y ) ∈ [ − 1 , 1 ] . cos(2y) \in [-1, 1]. Therefore x = 0 x = 0 is not contained anywhere on this curve and no tangents can be described there.

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