What is the general solution for 26^2=13^2+19^2-(13)(19)cos(J) ?

The general solution for 26^2=13^2+19^2-(13)(19)cos(J) is J=2.20320…+2pin,J=-2.20320…+2pin

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26^2=13^2+19^2-(13)(19)cos(J)

2 6^{2} = 1 3^{2} + 1 9^{2} - (13) (19) cos (J)

J = 2.20320 \dots + 2 πn, J = - 2.20320 \dots + 2 πn

Solution steps

2 6^{2} = 1 3^{2} + 1 9^{2} - (13) (19) cos (J)

Switch sides

1 3^{2} + 1 9^{2} - 13 \cdot 19 cos (J) = 2 6^{2}

Multiply the numbers:

13 \cdot 19 = 247

1 3^{2} + 1 9^{2} - 247 cos (J) = 2 6^{2}

1 3^{2} = 169

169 + 1 9^{2} - 247 cos (J) = 2 6^{2}

1 9^{2} = 361

169 + 361 - 247 cos (J) = 2 6^{2}

Add the numbers:

169 + 361 = 530

- 247 cos (J) + 530 = 2 6^{2}

2 6^{2} = 676

- 247 cos (J) + 530 = 676

Move

530

to the right side

- 247 cos (J) = 146

Divide both sides by

- 247

cos (J) = - \frac{146}{247}

Apply trig inverse properties

J = arccos (- \frac{146}{247}) + 2 πn, J = - arccos (- \frac{146}{247}) + 2 πn

Show solutions in decimal form

J = 2.20320 \dots + 2 πn, J = - 2.20320 \dots + 2 πn

4 = sec^{2} (x)

tan (\frac{x}{2}) = 4

sec (x) sin (x) + sec (x) = 2 sec (x)

3 - sin (x) (4 + 6 sin (x)) = 0

cos (a) = - \frac{3}{5}, sin (a) > 0

What is the general solution for 26^2=13^2+19^2-(13)(19)cos(J) ?
The general solution for 26^2=13^2+19^2-(13)(19)cos(J) is J=2.20320…+2pin,J=-2.20320…+2pin