Question

4 sin ^2x-4 sin x+1 =0

Answer 1

$4 sin ^2x-4 sin x+1 =0 \\(2sinx-1)^2=0 \\2sinx-1=0 \\2sinx=1 \\sinx= \frac{1}{2} \\x_1= \frac{\pi}{6} +2\pi n,\ n \in Z \\x_2=\frac{5\pi}{6} +2\pi n,\ n \in Z$

Answer 2

To solve the equation 4sin^2(x) - 4sin(x) + 1 = 0, we can use the quadratic formula. Let's consider the equation in terms of sin(x):

4sin^2(x) - 4sin(x) + 1 = 0

Now, let's compare it with the standard form of a quadratic equation:

ax^2 + bx + c = 0

In our case, a = 4, b = -4, and c = 1. Now, we can use the quadratic formula to find the value(s) of sin(x):

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values from our equation:

sin(x) = [4 ± √((-4)^2 - 4 * 4 * 1)] / (2 * 4) sin(x) = [4 ± √(16 - 16)] / 8 sin(x) = (4 ± √0) / 8

Now, since the discriminant (b^2 - 4ac) is zero, we will get a single real solution for sin(x):

sin(x) = (4 + 0) / 8 sin(x) = 4 / 8 sin(x) = 1/2

So, the solution for the equation is sin(x) = 1/2.

Now, to find the values of x, we can take the inverse sine (arcsin) of both sides:

x = arcsin(1/2)

The arcsin function gives us the angle whose sine is 1/2. There are two solutions within the range of [-π/2, π/2] that satisfy sin(x) = 1/2, which are:

1. x = π/6 (30 degrees)
2. x = 5π/6 (150 degrees)

So, the solutions to the equation 4sin^2(x) - 4sin(x) + 1 = 0 are x = π/6 and x = 5π/6 (in radians) or x = 30 degrees and x = 150 degrees.