A) 2
B) 3
C) 4
D) (d) 5
Correct Answer: C
Solution :
We have, \[2{{\sin }^{3}}x+2{{\cos }^{3}}x-3\sin 2x+2=0\] \[\Rightarrow \]\[2{{\sin }^{3}}x+2{{\cos }^{3}}x-3(2\sin x\cos x)+2=0\] \[\Rightarrow \]\[{{\sin }^{3}}x+{{\cos }^{3}}x-3\sin x\cos x+1=0\] \[\Rightarrow \] \[\sin x+\cos x+1=0\] [if \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0\]then \[a+b+c=0\]] \[\Rightarrow \] \[2\sin \frac{x}{2}\cos \frac{x}{2}+2{{\cos }^{2}}\frac{x}{2}=0\] \[\Rightarrow \] \[2\cos \frac{x}{2}\left( \sin \frac{x}{2}+\cos \frac{x}{2} \right)=0\] \[\Rightarrow \] \[2\cos \frac{x}{2}=0\,\,\,or\,\,\,\cos \frac{x}{2}+\sin \frac{x}{2}=0\] \[\Rightarrow \] \[\cos \frac{x}{2}=0\,\,\,or\,\,\tan \frac{x}{2}=-1\] If \[\cos \frac{x}{2}=0\] then \[\frac{x}{2}=\frac{\pi }{2},\frac{3\pi }{2}\] \[\Rightarrow \] \[x=\pi ,3\pi \] And if \[\tan \frac{x}{2}=-1,then\] \[\frac{x}{2}=\frac{3\pi }{4},\frac{7\pi }{4}\] \[\Rightarrow \] \[x=\frac{3\pi }{2},\frac{7\pi }{2}\] Hence, there are 4 solutions.
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