If 9sin2x-9cos2x-6 and 0 - x -2 then x -

The correct option is B π/4 3^2sin^2x+3^2cos^2x=3(2)=3+3 Therefore #160; 2sin^2x=2cos^2x=1 x=π4, π4,3π4,5π4 .However 0≤ x≤π2 Hence #1 ...

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Standard XII

Mathematics

Geometric Interpretation of Def.Int as Limit of Sum

If 9sin2x+9...

Question

If $9^{{sin}^{2} x} + 9^{{cos}^{2} x} = 6$ and $0 \leq x \leq \frac{π}{2}$ then x $=$

\frac{π}{6}, \frac{π}{3}

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\frac{π}{4}

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\frac{3 π}{4}, \frac{2 π}{3}

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\frac{π}{2}, \frac{4 π}{3}

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Solution

The correct option is B $\frac{π}{4}$
$3^{2 s i n^{2} x} + 3^{2 c o s^{2} x} = 3 (2) = 3 + 3$
Therefore
$2 s i n^{2} x = 2 c o s^{2} x = 1$
$x = \frac{π}{4}, \frac{- π}{4}, \frac{3 π}{4}, \frac{5 π}{4}$ .
However
$0 \leq x \leq \frac{π}{2}$
Hence
$x = \frac{π}{4}$

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If 9sin2x+9cos2x=6 and 0≤x≤π2 then x=

The correct option is B π432sin2x+32cos2x=3(2)=3+3Therefore 2sin2x=2cos2x=1x=π4,−π4,3π4,5π4.However0≤x≤π2Hence x=π4

If $9^{{sin}^{2} x} + 9^{{cos}^{2} x} = 6$ and $0 \leq x \leq \frac{π}{2}$ then x $=$

The correct option is B $\frac{π}{4}$
$3^{2 s i n^{2} x} + 3^{2 c o s^{2} x} = 3 (2) = 3 + 3$
Therefore
$2 s i n^{2} x = 2 c o s^{2} x = 1$
$x = \frac{π}{4}, \frac{- π}{4}, \frac{3 π}{4}, \frac{5 π}{4}$ .
However
$0 \leq x \leq \frac{π}{2}$
Hence
$x = \frac{π}{4}$