A) \[-\frac{7}{9}\]
B) \[-\frac{3}{5}\]
C) \[\frac{1}{3}\]
D) \[\frac{2}{9}\]
Correct Answer: A
Solution :
\[5\left[ \frac{1-t}{t}-t \right]=2(2t-1)+9\] \[\{\text{Let}\,{{\cos }^{2}}x=t\}\] \[\Rightarrow \]\[5(1-t-{{t}^{2}})=t(4t+7)\] \[\Rightarrow \]\[9{{t}^{2}}+12t-5=0\] \[\Rightarrow \]\[9{{t}^{2}}+15t-3t-5=0\] \[\Rightarrow \]\[(3t-1)(3t+5)=0\] \[\Rightarrow \]\[t=\frac{1}{3}\,\,as\,t\ne -\frac{5}{3}.\] \[\cos 2x=2\left( \frac{1}{3} \right)-1=-\frac{1}{3}\] \[\cos 4x=2{{\left( -\frac{1}{3} \right)}^{2}}-1=-\frac{7}{9}\]
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