A) one real solution
B) no real solution
C) more than one real solution
D) none of the above
Correct Answer: B
Solution :
\[{{x}^{2}}{{\cos }^{2}}x={{x}^{2}}{{\sin }^{2}}x+{{\sin }^{2}}x\] \[\Rightarrow \] \[{{x}^{2}}({{\cos }^{2}}x-{{\sin }^{2}}x)={{\sin }^{2}}x\] \[\Rightarrow \] \[{{x}^{2}}\cos 2x={{\sin }^{2}}x\] \[\Rightarrow \] \[{{x}^{2}}\cos 2x=\frac{1-\cos 2x}{2}\] \[\Rightarrow \] \[2{{x}^{2}}\cos 2x=1-\cos 2x\] \[\Rightarrow \] \[\cos 2x(2{{x}^{2}}+1)=1\] \[\Rightarrow \] \[\cos 2x=\frac{1}{2{{x}^{2}}+1}\] \[\because \] \[-1\le \cos 2x\le 1\] \[\therefore \] \[-1\le \frac{1}{2{{x}^{2}}+1}\le 1\] \[\Rightarrow \] \[\frac{1}{2{{x}^{2}}+1}\ge -1\] ? (i) and \[\frac{1}{2{{x}^{2}}+1}\le 1\] ? (ii) On solving Eqs. (i) and (ii), we get \[x=0\] But \[x=0\] does not satisfy the given equation, therefore no real solution exist.
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