Question 11 of 33MediumMCQ
The number of roots of the equation, (81)sin2x + (81)cos2x = 30 in the interval [ 0, \[\pi\] ] is equal to :
Solutions · 1
Official solution
\[{(81)^{{{\sin }^2}x}} + {(81)^{1 - {{\sin }^2}x}} = 30\]
\[{(81)^{{{\sin }^2}x}} + {{81} \over {{{(81)}^{{{\sin }^2}x}}}} = 30\]
Let \[{(81)^{{{\sin }^2}x}} = t\]
\[t + {{81} \over t} = 30 \Rightarrow {t^2} + 81 = 30t\]
t2−30t+81=0
t2−27t−3t+81=0
(t−3)(t−27)=0
t=3,27
\[{(81)^{{{\sin }^2}x}} = 3,{3^3}\]
\[{3^{4{{\sin }^2}x}} = {3^1},{3^3}\]
\[4{\sin ^2}x = 1,3\]
\[{\sin ^2}x = {1 \over 4},{3 \over 4}\]
in [0, \[\pi\] ] sin x > 0
\[\sin x = {1 \over 2},{{\sqrt 3 } \over 2}\]
\[x = {\pi \over 6},{{5\pi } \over 6},{\pi \over 3},{{2\pi } \over 3}\]
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