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2 years ago

What are the zeros of f(x)= 2 cos (3x) +1 on the interval [0,pi]

Mathematics

1 answer:

Hitman42 [59]2 years ago

7 0

Answer:

40° or 2π/9 rad

Step-by-step explanation:

Let f(x) = 0 and solve

2cos (3x) + 1 = 0

2cos (3x) = -1

cos (3x) = -1/2

3x = arccos (-1/2) = 120

x = 40° or 2π/9 rad

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3 years ago

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<u>Answer:</u>

The equation of a polynomial of degree 3, with zeros 1, 2 and -1 is $x^{3}-2 x^{2}-x+2=0$

<u>Solution:</u>

Given, the polynomial has degree 3 and roots as 1, 2, and -1. And f(0) = 2.

We have to find the equation of the above polynomial.

We know that, general equation of 3rd degree polynomial is

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Here in our problem, a = 1, b = 2, c = -1.

Substitute the above values in f(x)

$F(x)=x^{3}-(1+2+(-1)) x^{2}+(1 \times 2+2(-1)+1(-1)) x-1 \times 2 \times(-1)=0$

$\begin{array}{l}{\rightarrow x^{3}-(3-1) x^{2}+(2-2-1) x-(-2)=0} \\ {\rightarrow x^{3}-(2) x^{2}+(-1) x-(-2)=0} \\ {\rightarrow x^{3}-2 x^{2}-x+2=0}\end{array}$

So, the equation is $x^{3}-2 x^{2}-x+2=0$

Let us put x = 0 in f(x) to check whether our answer is correct or not.

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