What is the general solution for cos(A)= 1/(sqrt(65)),cos(B)= 1/(sqrt(5)) ?

The general solution for cos(A)= 1/(sqrt(65)),cos(B)= 1/(sqrt(5)) is No Solution for A\in\mathbb{R}

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cos(A)= 1/(sqrt(65)),cos(B)= 1/(sqrt(5))

cos (A) = \frac{1}{65}, cos (B) = \frac{1}{5}

No Solution for A \in R

Solution steps

cos (A) = \frac{1}{65}, cos (B) = \frac{1}{5}

Simplify

\frac{1}{65} : \frac{65}{65}

\frac{1}{65}

Multiply by the conjugate

\frac{65}{65}

= \frac{1 \cdot 65}{65 65}

1 \cdot 65 = 65

6565 = 65

6565

Apply radical rule:

a a = a

6565 = 65

= 65

= \frac{65}{65}

cos (A) = \frac{65}{65}

Apply trig inverse properties

cos (A) = \frac{65}{65}

General solutions for

cos (A) = \frac{65}{65}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

A = arccos (\frac{65}{65}) + 2 πn, A = 2 π - arccos (\frac{65}{65}) + 2 πn

A = arccos (\frac{65}{65}) + 2 πn, A = 2 π - arccos (\frac{65}{65}) + 2 πn

Solutions for the range

cos (B) = \frac{1}{5}

No Solution for A \in R

sin (θ) = 0.9063

a \cdot sin (x) + b \cdot cos (x) = 0

cos (θ) = - 0.19

9 cos^{2} (x) = 2

tan (θ) = \frac{6}{14}

What is the general solution for cos(A)= 1/(sqrt(65)),cos(B)= 1/(sqrt(5)) ?
The general solution for cos(A)= 1/(sqrt(65)),cos(B)= 1/(sqrt(5)) is No Solution for A\in\mathbb{R}