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Find all the solutions of the equation in the interval [0, 2π). 4sin(x) = -cos^2(x).

a) No solution
b) π/6, 5π/6
c) π/3, 2π/3
d) π/4, 3π/4

1 Answer

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Final answer:

The quadratic equation in sine derived from 4sin(x) = -cos^2(x) has no real solutions, therefore, there are no solutions to the equation within the interval [0, 2π).

Step-by-step explanation:

The equation we're looking at is 4sin(x) = -cos^2(x). To solve this equation, we can use the Pythagorean identity which states that sin^2(x) + cos^2(x) = 1. Substituting for cos^2(x), we have 4sin(x) = -(1 - sin^2(x)). This simplifies to sin^2(x) + 4sin(x) + 1 = 0.

This is a quadratic equation in terms of sin(x), which we can solve using the quadratic formula.

Unfortunately, when we apply the formula, we find that the discriminant (b^2 - 4ac) is negative, indicating there are no real solutions to this equation.

Therefore, in the interval [0, 2π), there are no solutions to the equation 4sin(x) = -cos^2(x).

User Allen Zeng
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