How do you solve 1+sinx=2cos^2x in the interval 0<=x<=2pi?

1 Answer
sudheer22kumar
Jul 26, 2016

Based on two different cases: x = pi/6, (5pi)/6 or (3pi)/2

Look below for the explanation of these two cases.

Explanation:

Since, cos^ x + sin^2 x = 1
we have: cos^2 x = 1 - sin^2 x

So we can replace cos^2 x in the equation 1 + sinx = 2cos^2x by (1- sin^2 x)

=>2 (1 - sin^2 x) = sin x +1

or, 2 - 2 sin^2 x = sin x + 1

or, 0=2sin ^2 x + sin x + 1 - 2

or, 2sin ^2 x + sin x - 1 = 0

using the quadratic formula:

x = (-b+-sqrt(b^2 - 4ac))/(2a) for quadratic equation ax^2+bx+c=0

we have:

sin x = (-1+-sqrt(1^2 - 4*2*(-1)))/(2*2)

or, sin x = (-1+-sqrt(1 + 8))/4

or, sin x = (-1+-sqrt(9))/4

or, sin x = (-1+-3)/4

or, sin x = (-1+3)/4, (-1-3)/4

or, sin x = 1/2, -1

Case I:

sin x = 1/2

for the condition: 0<=x<=2pi

we have:

x= pi/6 or (5pi)/6 to get positive value of sinx

Case II:

sin x = -1

we have:

x= (3pi)/2 to get negative value of sinx

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