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(A) The given equation is $2 	heta - \cos^{2} 	heta + \sqrt{2} = 0$.This can be rewritten as $\cos^{2} 	heta = 2 	heta + \sqrt{2}$.Let $f(	heta)

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$1$

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(A) The given equation is $2 	heta - \cos^{2} 	heta + \sqrt{2} = 0$.This can be rewritten as $\cos^{2} 	heta = 2 	heta + \sqrt{2}$.Let $f(	heta) = \cos^{2} 	heta$ and $g(	heta) = 2 	heta + \sqrt{2}$.The function $f(	heta) = \cos^{2} 	heta$ is a periodic function with a range of $[0, 1]$.The function $g(	heta) = 2 	heta + \sqrt{2}$ is a straight line with a slope of $2$ and a $y$-intercept of $\sqrt{2} \approx 1.414$.Since the maximum value of $\cos^{2} 	heta$ is $1$,and for $	heta > 0$,$g(	heta) > \sqrt{2} > 1$,there are no solutions for $	heta > 0$.For $	heta < 0$,the line $g(	heta)$ intersects the curve $f(	heta)$ at exactly one point,as shown in the graph.

The number of solutions of the equation $2 \theta - \cos^{2} \theta + \sqrt{2} = 0$ is equal to

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