NCERT Problem
Source:-Miscellaneous Exercise on Chapter 2 (12th class) If:- sin − 1 ( 1 − x ) − 2 s i n − 1 x = 2 π Then find the vale of x
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2 solutions
Oh made a mistake got the answer but did'nt checked the extraneous root
s i n − 1 ( 1 − x ) − 2 π = 2 s i n − 1 x
c o s ( s i n − 1 ( 1 − x ) ) − 2 π = c o s ( 2 s i n − 1 x )
c o s 2 x = 2 s i n 2 x − 1
s i n ( s i n − 1 ( 1 − x ) ) = 1 − 2 s i n 2 ( s i n − 1 x )
⟹ 1 − x = 1 − 2 x 2
2 x 2 − x = 0
x ( 2 x − 1 ) = 0
⟹ x = 0 o r x = 1 / 2
only 0 satisfy the given expression(I did'nt checked this)
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X=1÷2 also satisfy it as sin^(-1)0.5 can be 30° and 150°.So it can be 150°-2*30°=90°.
I also made the same mistake when i was solving this question
arcsin(1-x) - 2 (arcsin (x)) = (pi/2) arcsin (1-x) = (pi/2) + 2 (arcsin (x)) (1-x) = sin ((pi/2) + 2 (arcsin(x)) = cos(2 arcsin (x)) = 1 - 2( sin(arcsin (x)))^2 = 1 - 2x^2 x - 2x^2 = 0 x(1 - 2x) = 0 => x = 0 or (1/2) Substituting values of x into equation, only x = 0 is applicable.