Trigonometry with quadratic

Geometry Level 3

The values of a and b for which the quadratic equation x 2 + 2 x + 2 + e − 2 a − c o s b = 0 x^{2}+2x+2+e^{-2a}-cosb=0 has a real solution is

no real values of a and b a=b=1 a€(0,1) b€(90,360) a,b €R

Harsh Poddar by harsh poddar , 22, India

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1 solution

Chitrang Garg
Jul 28, 2015

the quadratic in x (ie. x^2 + 2x + 2) has the minimum possible value equal to 1. Any exponent to 'e' cannot give a value less than or equal to 0. Hence, to get a real solution, cosb needs to be greater that 1 (which of course is never possible).

Thus, no possible real solution.

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