The value-s of x -0- -2 satisfying -3-1-sin x-3-1-cos x-4 -2 areA- -12B- 12-7 -D- 14-36

The correct options are A π12 D 11π36√(3) 12√(2)sin x+ √(3)+12√(2)cos x = 2 sinπ12cos x+ cosπ12sin x = sin 2x sin 2x = sin ( x+ π12 ) ∴ 2x = x + π12 or 2x= π ...

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Standard XII

Mathematics

Principal Solution of Trigonometric Equation

The value's o...

Question

The value's of $x \in (0, \frac{π}{2})$ satisfying $\frac{\sqrt{3} - 1}{sin x} + \frac{\sqrt{3} + 1}{cos x} = 4 \sqrt{2}$ are

\frac{π}{12}

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\frac{5 π}{12}

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\frac{7 π}{24}

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\frac{11 π}{36}

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Solution

The correct options are
A $\frac{π}{12}$
D $\frac{11 π}{36}$
$\frac{\sqrt{3} - 1}{2 \sqrt{2} sin x} + \frac{\sqrt{3} + 1}{2 \sqrt{2} cos x} = 2$
$sin \frac{π}{12} cos x + cos \frac{π}{12} sin x = sin 2 x$
$sin 2 x = sin (x + \frac{π}{12})$
$∴ 2 x = x + \frac{π}{12} or 2 x = π - (x + \frac{π}{12}) (∵ 2 x \in (0, π))$
$x = \frac{π}{12} or 3 x = \frac{11 π}{12}$
$∴ x = \frac{π}{12} or \frac{11 π}{36}$

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Principal Solution of Trigonometric Equation

Standard XII Mathematics

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The value's of x∈(0,π2) satisfying √3−1sinx+√3+1cosx=4√2 are

The correct options are A π12 D 11π36√3−12√2sinx+√3+12√2cosx=2 sinπ12cosx+cosπ12sinx=sin2x sin2x=sin(x+π12) ∴2x=x+π12 or 2x=π−(x+π12) (∵2x∈(0,π)) x=π12 or 3x=11π12 ∴x=π12 or 11π36

The value's of $x \in (0, \frac{π}{2})$ satisfying $\frac{\sqrt{3} - 1}{sin x} + \frac{\sqrt{3} + 1}{cos x} = 4 \sqrt{2}$ are