solve: 2sin^2x=1

Question

Solve each equation on the interval [0𝜋, 2𝜋).

Answer 1

2sin²(x)=1 -> sin²(x)=1/2 -> sin(x)=±√2/2 -> x=π/4, 3π/4, 5π/4, 7π/4

Answer 2

The equation is 2*(sin(x))^2 = 1, and we need to find a value of x that makes this equality true.

So, we can do some algebra. For example, divide both sides by two:

sin(x)^2 = 1/2

Take the square root of both sides

sin(x) = +/- sqrt(1/2) = +/- 1/(sqrt(2))

Now, if we consult a unit circle, we can see that an angle of pi/4 or 3pi/4 has a sine of sqrt(2)/2,

And an angle of 5pi/4 or 7pi/4 has a sine of -sqrt(2)/2.

Algebra tells us that 1/(sqrt(2)) = sqrt(2)/2

Thus, x must be either pi/4 radians, 3pi/4 radians, 5pi/4 radians, or 7pi/4 radians.

As it tells us to find a value between 0 and 2 pi, any of these values of x is a possible solution