What is the general solution for 10=sqrt(65)*sqrt(5)*cos(θ) ?

The general solution for 10=sqrt(65)*sqrt(5)*cos(θ) is θ=0.98279…+2pin,θ=2pi-0.98279…+2pin

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10=sqrt(65)sqrt(5)cos(θ)

10 = 65 \cdot 5 \cdot cos (θ)

θ = 0.98279 \dots + 2 πn, θ = 2 π - 0.98279 \dots + 2 πn

Degrees

θ = 56.30993 \dots^{\circ} + 36 0^{\circ} n, θ = 303.69006 \dots^{\circ} + 36 0^{\circ} n

Solution steps

10 = 655 cos (θ)

Switch sides

655 cos (θ) = 10

Simplify

655 : 513

655

Factor integer

65 = 5 \cdot 13

= 5 \cdot 13 5

Apply radical rule:

5 \cdot 13 = 513

= 5135

Apply radical rule:

a a = a

55 = 5

= 513

513 cos (θ) = 10

Divide both sides by

513

513 cos (θ) = 10

Divide both sides by

513

\frac{5 13 cos ( θ )}{5 13} = \frac{10}{5 13}

Simplify

\frac{5 13 cos ( θ )}{5 13} = \frac{10}{5 13}

Simplify

\frac{5 13 cos ( θ )}{5 13} : cos (θ)

\frac{5 13 cos ( θ )}{5 13}

Divide the numbers:

\frac{5}{5} = 1

= \frac{13 cos ( θ )}{13}

Cancel the common factor:

13

= cos (θ)

Simplify

\frac{10}{5 13} : \frac{2 13}{13}

\frac{10}{5 13}

Divide the numbers:

\frac{10}{5} = 2

= \frac{2}{13}

Rationalize

\frac{2}{13} : \frac{2 13}{13}

\frac{2}{13}

Multiply by the conjugate

\frac{13}{13}

= \frac{2 13}{13 13}

1313 = 13

1313

Apply radical rule:

a a = a

1313 = 13

= 13

= \frac{2 13}{13}

= \frac{2 13}{13}

cos (θ) = \frac{2 13}{13}

cos (θ) = \frac{2 13}{13}

cos (θ) = \frac{2 13}{13}

Apply trig inverse properties

cos (θ) = \frac{2 13}{13}

General solutions for

cos (θ) = \frac{2 13}{13}

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

θ = arccos (\frac{2 13}{13}) + 2 πn, θ = 2 π - arccos (\frac{2 13}{13}) + 2 πn

θ = arccos (\frac{2 13}{13}) + 2 πn, θ = 2 π - arccos (\frac{2 13}{13}) + 2 πn

Show solutions in decimal form

θ = 0.98279 \dots + 2 πn, θ = 2 π - 0.98279 \dots + 2 πn

What is the general solution for 10=sqrt(65)*sqrt(5)*cos(θ) ?
The general solution for 10=sqrt(65)*sqrt(5)*cos(θ) is θ=0.98279…+2pin,θ=2pi-0.98279…+2pin