What is the general solution for 40^2=28^2+15^2-2(28)(15)*cos(x) ?

The general solution for 40^2=28^2+15^2-2(28)(15)*cos(x) is x=2.35120…+2pin,x=-2.35120…+2pin

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40^2=28^2+15^2-2(28)(15)*cos(x)

4 0^{2} = 2 8^{2} + 1 5^{2} - 2 (28) (15) \cdot cos (x)

x = 2.35120 \dots + 2 πn, x = - 2.35120 \dots + 2 πn

Solution steps

4 0^{2} = 2 8^{2} + 1 5^{2} - 2 (28) (15) cos (x)

Switch sides

2 8^{2} + 1 5^{2} - 2 \cdot 28 \cdot 15 cos (x) = 4 0^{2}

Multiply the numbers:

2 \cdot 28 \cdot 15 = 840

2 8^{2} + 1 5^{2} - 840 cos (x) = 4 0^{2}

2 8^{2} = 784

784 + 1 5^{2} - 840 cos (x) = 4 0^{2}

1 5^{2} = 225

784 + 225 - 840 cos (x) = 4 0^{2}

Add the numbers:

784 + 225 = 1009

- 840 cos (x) + 1009 = 4 0^{2}

4 0^{2} = 1600

- 840 cos (x) + 1009 = 1600

Move

1009

to the right side

- 840 cos (x) = 591

Divide both sides by

- 840

cos (x) = - \frac{197}{280}

Apply trig inverse properties

x = arccos (- \frac{197}{280}) + 2 πn, x = - arccos (- \frac{197}{280}) + 2 πn

Show solutions in decimal form

x = 2.35120 \dots + 2 πn, x = - 2.35120 \dots + 2 πn

5 tan^{2} (x) - 1 = 2 tan^{2} (x)

5 sin (2 x) + 8 cos (x) = 0

cos^{2} (x) tan^{2} (x) csc^{2} (x) = sin (x)

cos (a) - sin (a) = cot (a) cos (a)

15 cos (\frac{2 π}{365} t) + 43 = 30

What is the general solution for 40^2=28^2+15^2-2(28)(15)*cos(x) ?
The general solution for 40^2=28^2+15^2-2(28)(15)*cos(x) is x=2.35120…+2pin,x=-2.35120…+2pin