What is the general solution for cos(A)=(4^2+(5.1)^2-3^2)/(2-4*5.1) ?

The general solution for cos(A)=(4^2+(5.1)^2-3^2)/(2-4*5.1) is No Solution for A\in\mathbb{R}

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cos(A)=(4^2+(5.1)^2-3^2)/(2-4*5.1)

cos (A) = \frac{4 ^{2} + ( 5.1 ) ^{2} - 3 ^{2}}{2 - 4 \cdot 5.1}

No Solution for A \in R

Solution steps

cos (A) = \frac{4 ^{2} + ( 5.1 ) ^{2} - 3 ^{2}}{2 - 4 \cdot 5.1}

Simplify

\frac{4 ^{2} + 5. 1 ^{2} - 3 ^{2}}{2 - 4 \cdot 5.1} : - \frac{33.01}{18.4}

\frac{4 ^{2} + 5. 1 ^{2} - 3 ^{2}}{2 - 4 \cdot 5.1}

4^{2} + 5. 1^{2} - 3^{2} = 33.01

4^{2} + 5. 1^{2} - 3^{2}

4^{2} = 16

= 16 + 5. 1^{2} - 3^{2}

5. 1^{2} = 26.01

= 16 + 26.01 - 3^{2}

3^{2} = 9

= 16 + 26.01 - 9

Add/Subtract the numbers:

16 + 26.01 - 9 = 33.01

= 33.01

= \frac{33.01}{2 - 4 \cdot 5.1}

2 - 4 \cdot 5.1 = - 18.4

2 - 4 \cdot 5.1

Multiply the numbers:

4 \cdot 5.1 = 20.4

= 2 - 20.4

Subtract the numbers:

2 - 20.4 = - 18.4

= - 18.4

= \frac{33.01}{- 18.4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{33.01}{18.4}

- 1 \leq cos (x) \leq 1

No Solution for A \in R

cos (a) = \frac{5}{4}

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

2 cos (2 θ + 1 0^{\circ}) = 1

cot (β) = \frac{- 3}{3}

2 sin^{2} (3 x) - 3 cos (3 x) + 1 = 0

What is the general solution for cos(A)=(4^2+(5.1)^2-3^2)/(2-4*5.1) ?
The general solution for cos(A)=(4^2+(5.1)^2-3^2)/(2-4*5.1) is No Solution for A\in\mathbb{R}