Question

Algebraically solve the exact values of all angles in the interval [0,pi] that satisfy the equation: 2sin^2(x) = 1 - cos(x)

Answer 1

Given equation is

`2\sin ^2(x)=1-\cos (x)`
`\text{ Use }\sin ^2(x)=1-\cos ^2(x).`

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

`2(1^2-\cos ^2(x))=1-\cos (x)`
`\text{ Use }1^2-\cos ^2(x)=(1-\cos (x))(1+\cos (x)).`
`2(1-\cos (x))(1+\cos (x))=1-\cos (x)`

Cancel out the common term on both sides, we get

`2(1+\cos (x))=1`

Dividing both sides by 2, we get

`1+\cos (x)=(1)/(2)`

Subtracting 1 from both sides, we get

`1+\cos (x)-1=(1)/(2)-1`

`\cos (x)=-(1)/(2)`

`\text{Use }\cos ((2\pi)/(3))=-(1)/(2)\text{.}`

`\cos (x)=\cos ((2\pi)/(3))`
`x=(2\pi)/(3)`

Hence the exact value of x is

`x=(2\pi)/(3)`

The exact values of all angles in the interval [0, pi] that satisfy the given equation is 2pi/3.