# If asked to solve the following equation: `2cos^2(x) = 1 + sin(x)` by first finding the zeroes (x-values) - doing this by manipulating the equation using trigonometric identities, and other...

If asked to solve the following equation:

`2cos^2(x) = 1 + sin(x)`

by first finding the zeroes (x-values) - doing this by manipulating the equation using trigonometric identities, and other algebraic methods, are the following solutions correct for `0<=x<=2pi` :

`Pi/3 , (5Pi)/(3) , 2Pi`

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### 2 Answers

`2cos^2x = (1+sinx)`

`sin^2x+cos^2x = 1`

`cos^2x = 1-sin^2x`

`1-sin^2x = (1-sinx)(1+sinx)`

`2cos^2x = (1+sinx)`

`2(1-sinx)(1+sinx) = (1+sinx)`

`2(1+sinx)(1-sinx)-(1+sinx) = 0`

`(1+sinx)(2-2sinx-1) = 0`

`(1+sinx)(1-2sinx) = 0`

`1+sinx = 0`

`sinx = -1`

`sinx = sin(-pi/2)`

`x = npi+(-1)^n(-pi/2)` where `n in Z`

`(n,x)`

`(0,-pi/2)`

`(1,pi/2)`

`(2,(3pi)/2)`

`(3,(7pi)/2)`

`(1-2sinx) = 0`

`sinx = 1/2`

`sinx = sin(pi/6)`

`x = mpi+(-1)^m(pi/6) ` where `m in Z`

`(m,x)`

`(0,pi/6)`

`(1,(5pi)/6)`

`(2,(13pi)/6)`

`(3,(17pi)/6)`

*So the answer are;*

`x = pi/6`

`x = pi/2`

`x = (3pi)/2`

`x = (5pi)/6`

**Sources:**

`cos^2(x) = (1-sin^2(x))`

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

`2 - 2sin^2(x) - 1 - sin(x) = 0`

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

Using your knowledge of quatratic expressions, we can factor this, as it's in the form of ax^2 + bx + c,

`-(2sin(x) + 1)(sin(x) - 1) = 0` So,

`sin(x) = 1/2`

and

`sin(x) = -1`

Using the unit circle or trig identities, what values of x makes this true?

For `0 lt x lt 2pi`

`sin(x) = 1/2` when `x = pi/6, x = 5pi/6`

`sin(x) = -1` when `x = (3pi)/2`