Find the angle a for the identity 2sin^2a-sina-1=0 to be true.
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We have to find a for the identity 2sin^2a-sina-1=0 to be true.
Now 2* (sin a)^2 - sin a - 1 =0
let y = sin a
=> 2y^2 - y - 1 =0
=> 2y^2 - 2y + y - 1 =0
=> 2y( y - 1) + 1(y-1) = 0
=> (2y + 1)( y-1) = 0
=> y = -1/2 and y = 1
As y = sin x
x = arc sin (-1/2)
=>x = -30 + k*360 degrees
x = arc sin 1
=> x = 90 + k*360 degrees
Therefore x is -30 + k*360 degrees and 90 + k*360 degrees.
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We'll apply substitution technique and we'll note sin a = t.
We'll re-write the equation using the new variable t:
2t^2 - t - 1 = 0
Since it is a quadratic equation, we'll apply the quadratic formula:
t1 = [1 + sqrt(1 + 8)]/4
t1 = (1+3)/4
t1 = 1
t2 = (1-3)/4
t2 = -1/2
We'll put sin a = t1.
sin a = 1
a = (-1)^k*arc sin 1 + k*pi
a = (-1)^k*(pi/2) + k*pi
Now, we'll put sin a = t2.
sin a = -1/2
a = (-1)^k*arcsin(-1/2) + k*pi
a = (-1)^(k+1)*arcsin(1/2) + k*pi
a = (-1)^(k+1)*(pi/6) + k*pi
The solutions of the equation are the values of measures of the angle a: {(-1)^k*(pi/2) + k*pi}U{(-1)^(k+1)*(pi/6) + k*pi}.
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