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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Popular Trigonometry >

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

Solution

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

Solution

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

Degrees

x = 134.71424 \dots^{\circ} + 36 0^{\circ} n, x = - 134.71424 \dots^{\circ} + 36 0^{\circ} n

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) + 1009 = 1600

Subtract

1009

from both sides

- 840 cos (x) + 1009 - 1009 = 1600 - 1009

Simplify

- 840 cos (x) = 591

- 840 cos (x) = 591

Divide both sides by

- 840

- 840 cos (x) = 591

Divide both sides by

- 840

\frac{- 840 cos ( x )}{- 840} = \frac{591}{- 840}

Simplify

\frac{- 840 cos ( x )}{- 840} = \frac{591}{- 840}

Simplify

\frac{- 840 cos ( x )}{- 840} : cos (x)

\frac{- 840 cos ( x )}{- 840}

Apply the fraction rule:

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

= \frac{840 cos ( x )}{840}

Divide the numbers:

\frac{840}{840} = 1

= cos (x)

Simplify

\frac{591}{- 840} : - \frac{197}{280}

\frac{591}{- 840}

Apply the fraction rule:

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

= - \frac{591}{840}

Cancel the common factor:

3

= - \frac{197}{280}

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Graph

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

15cos((2pi)/(365)t)+43=30

15 cos (\frac{2 π}{365} t) + 43 = 30

tan^2(x)=2sec(x)-1

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

cos(z)=1

cos (z) = 1

cos(z)=5

cos (z) = 5

Frequently Asked Questions (FAQ)

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