What is the general solution for arctan(x^2-2x)=0 ?

The general solution for arctan(x^2-2x)=0 is x=2,x=0

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arctan(x^2-2x)=0

Solution

arctan (x^{2} - 2 x) = 0

Solution

x = 2, x = 0

Solution steps

arctan (x^{2} - 2 x) = 0

Apply trig inverse properties

arctan (x^{2} - 2 x) = 0

arctan (x) = a \Rightarrow x = tan (a)

x^{2} - 2 x = tan (0)

tan (0) = 0

tan (0)

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

= 0

x^{2} - 2 x = 0

x^{2} - 2 x = 0

Solve

x^{2} - 2 x = 0 : x = 2, x = 0

x^{2} - 2 x = 0

Solve with the quadratic formula

x^{2} - 2 x = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 0

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

x_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2}{2 \cdot 1} : 2

\frac{- ( - 2 ) + 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 + 2}{2 \cdot 1}

Add the numbers:

2 + 2 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

x = \frac{- ( - 2 ) - 2}{2 \cdot 1} : 0

\frac{- ( - 2 ) - 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 - 2}{2 \cdot 1}

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

x = 2, x = 0

x = 2, x = 0

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sin (x^{\circ}) = \frac{6.57 \cdot 1 0 ^{- 7}}{1.8 \cdot 1 0 ^{- 6}}

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3 sin^{2} (x) + cos^{2} (x) = 3 sin (x)

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(\frac{( 1 - sin ^{2} ( x ) )}{( 1 + 2 \cdot sin ^{2} ( x ) )}) = 0.45

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Frequently Asked Questions (FAQ)

What is the general solution for arctan(x^2-2x)=0 ?
The general solution for arctan(x^2-2x)=0 is x=2,x=0