# Determine the angle. Determine the angle a for the identity 2sin^2a-sina-1=0 to be true.

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You may use the alternative method to solve for a the equation, hence, you may use factorization, such that:

`sin^2 a + sin^2 a - sin a - 1 = 0`

You may form two groups of terms such that:

`(sin^2 a - sin a) + (sin^2 a - 1) = 0`

You may factor out `sin a` in the first group, such that:

`sin a(sin a - 1) + (sin^2 a - 1) = 0`

You may convert the difference of squares `sin^2 a - 1` in a product of two factors, such that:

`sin a(sin a - 1) + (sin a - 1)(sin a + 1) = 0`

Factoring out `sin a - 1` yields:

`(sin a - 1)(sin a + sin a + 1) = 0 => {(sin a - 1 = 0),(2sina + 1 = 0):}`

`sin a = 1 => a = (-1)^n*(pi/2) + n*pi , n in Z`

`sin a = -1/2 => a = (-1)^n(7pi/6) + n*pi , n in Z`

Hence, evaluating the solutions to the given trigonometric equation yields  `a = (-1)^n*(pi/2) + n*pi , n in Z , a = (-1)^n(7pi/6) + n*pi , n in Z.`

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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

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}.