The solution of the equation -sin x-cos x-1-sin 2 x-2- x - isA- -2B- -C- -4D- 3 -4E- -3

The correct option is C Let I = [sin x cos x]^1+sin 2x = [ √(2)sin ( π/4 + x ) ]^1+sin 2x At, x = π/4, I = [ √(2)sin ( π/4 + π/4 ) ]^1+sin2π/4 = (√(2))^2 ...

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Standard XI

Mathematics

Equations involving sinx + cosx and sinx.cosx

The solution ...

Question

The solution of the equation $[sin x + cos x]^{1 + sin 2 x} = 2, - π \leq x \leq π$ is

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Solution

The correct option is C
Let $I = [sin x cos x]^{1 + sin 2 x} = {[\sqrt{2} sin (\frac{π}{4} + x)]}^{1 + sin 2 x}$
At, $x = \frac{π}{4},$
$I = {[\sqrt{2} sin (\frac{π}{4} + \frac{π}{4})]}^{1 + sin \frac{2 π}{4}}$
$= (\sqrt{2})^{2} = 2$
Hence, option (c) is correct.

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The solution of the equation [sinx+cosx]1+sin2x=2,−π≤x≤π is

The correct option is C Let I=[sinxcosx]1+sin2x=[√2sin(π4+x)]1+sin2x At, x=π4, I=[√2sin(π4+π4)]1+sin2π4 =(√2)2=2 Hence, option (c) is correct.

The solution of the equation $[sin x + cos x]^{1 + sin 2 x} = 2, - π \leq x \leq π$ is

The correct option is C
Let $I = [sin x cos x]^{1 + sin 2 x} = {[\sqrt{2} sin (\frac{π}{4} + x)]}^{1 + sin 2 x}$
At, $x = \frac{π}{4},$
$I = {[\sqrt{2} sin (\frac{π}{4} + \frac{π}{4})]}^{1 + sin \frac{2 π}{4}}$
$= (\sqrt{2})^{2} = 2$
Hence, option (c) is correct.