If for real values of x,cos theta = x + frac{ | vedclass

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Question

(d) The quadratic equation is ${x^2} - x\cos 	heta + 1 = 0$

But $x$ is real, therefore ${B^2} - 4AC \ge 0$

$ \Rightarrow {\cos ^2}	heta \ge 4(

Vedclass · Accepted Answer

No value of $	heta $ is possible

Vedclass · Answer

(d) The quadratic equation is ${x^2} - x\cos 	heta + 1 = 0$

But $x$ is real, therefore ${B^2} - 4AC \ge 0$

$ \Rightarrow {\cos ^2}	heta \ge 4(1)(1) \Rightarrow {\cos ^2}	heta \ge 4$, which is impossible.

If for real values of $x,\cos \theta = x + \frac{1}{x},$ then

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