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# Find all solutions in the interval [0, 2π). 4 sin2x - 4 sin x + 1 = 0

**Solution:**

It is given that, equation:

4 sin^{2}x - 4 sin x + 1 = 0,

We can say sin x = a, then the equation becomes:

4a^{2} - 4a + 1 = 0

By dividing 4, we get,

a^{2} - a + 1/4

So, (a - 1/2) (a - 1/2) = 0

a = 1/2

Then,

sin x = 1/2

Because the interval between 0 - 2π and positive sin values are in quadrants 1 and 2 so the x values that meet are 30° and 150° or π/6 and 5/6 π.

Therefore, 30° and 150° or π/6 and 5/6 π are the solutions.

## Find all solutions in the interval [0, 2π). 4 sin2x - 4 sin x + 1 = 0

**Summary:**

All solutions in the interval [0, 2π) 4 sin^{2}x - 4 sin x + 1 = 0 are 30° and 150° or π/6 and 5/6 π.

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