What is the general solution for cos(3v)=cos(6),0<= v<= 360 ?

The general solution for cos(3v)=cos(6),0<= v<= 360 is No Solution for v\in\mathbb{R}

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cos(3v)=cos(6),0<= v<= 360

cos (3 v) = cos (6), 0^{\circ} \leq v^{\circ} \leq 36 0^{\circ}

No Solution for v \in R

Solution steps

cos (3 v) = cos (6), 0^{\circ} \leq v^{\circ} \leq 36 0^{\circ}

Apply trig inverse properties

cos (3 v) = cos (6)

General solutions for

cos (3 v) = cos (6)

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

3 v = arccos (cos (6)) + 36 0^{\circ} n, 3 v = 36 0^{\circ} - arccos (cos (6)) + 36 0^{\circ} n

3 v = arccos (cos (6)) + 36 0^{\circ} n, 3 v = 36 0^{\circ} - arccos (cos (6)) + 36 0^{\circ} n

Solve

3 v = arccos (cos (6)) + 36 0^{\circ} n : v = 12 0^{\circ} - 2 + \frac{36 0 ^{\circ} n}{3}

3 v = arccos (cos (6)) + 36 0^{\circ} n

Simplify

arccos (cos (6)) + 36 0^{\circ} n : 36 0^{\circ} - 6 + 36 0^{\circ} n

arccos (cos (6)) + 36 0^{\circ} n

arccos (cos (6)) = 36 0^{\circ} - 6

arccos (cos (6))

For

0 \leq v \leq 18 0^{\circ}, a rccos (cos (x)) = x

Rewrite using trig identities:

cos (6) = cos (36 0^{\circ} - 6)

cos (6)

cos (x) = cos (36 0^{\circ} - (x))

= cos (36 0^{\circ} - 6)

= arccos (cos (36 0^{\circ} - 6))

0 \leq 36 0^{\circ} - 6 \leq 18 0^{\circ}

= 36 0^{\circ} - 6

= 36 0^{\circ} - 6 + 36 0^{\circ} n

3 v = 36 0^{\circ} - 6 + 36 0^{\circ} n

Divide both sides by

3

3 v = 36 0^{\circ} - 6 + 36 0^{\circ} n

Divide both sides by

3

\frac{3 v}{3} = 12 0^{\circ} - \frac{6}{3} + \frac{36 0 ^{\circ} n}{3}

Simplify

v = 12 0^{\circ} - 2 + \frac{36 0 ^{\circ} n}{3}

v = 12 0^{\circ} - 2 + \frac{36 0 ^{\circ} n}{3}

Solve

3 v = 36 0^{\circ} - arccos (cos (6)) + 36 0^{\circ} n : v = 12 0^{\circ} - \frac{36 0 ^{\circ} - 6}{3} + \frac{36 0 ^{\circ} n}{3}

3 v = 36 0^{\circ} - arccos (cos (6)) + 36 0^{\circ} n

Simplify

36 0^{\circ} - arccos (cos (6)) + 36 0^{\circ} n : 36 0^{\circ} - (36 0^{\circ} - 6) + 36 0^{\circ} n

36 0^{\circ} - arccos (cos (6)) + 36 0^{\circ} n

arccos (cos (6)) = 36 0^{\circ} - 6

arccos (cos (6))

For

0 \leq v \leq 18 0^{\circ}, a rccos (cos (x)) = x

Rewrite using trig identities:

cos (6) = cos (36 0^{\circ} - 6)

cos (6)

cos (x) = cos (36 0^{\circ} - (x))

= cos (36 0^{\circ} - 6)

= arccos (cos (36 0^{\circ} - 6))

0 \leq 36 0^{\circ} - 6 \leq 18 0^{\circ}

= 36 0^{\circ} - 6

= 36 0^{\circ} - (36 0^{\circ} - 6) + 36 0^{\circ} n

3 v = 36 0^{\circ} - (36 0^{\circ} - 6) + 36 0^{\circ} n

Divide both sides by

3

3 v = 36 0^{\circ} - (36 0^{\circ} - 6) + 36 0^{\circ} n

Divide both sides by

3

\frac{3 v}{3} = 12 0^{\circ} - \frac{36 0 ^{\circ} - 6}{3} + \frac{36 0 ^{\circ} n}{3}

Simplify

v = 12 0^{\circ} - \frac{36 0 ^{\circ} - 6}{3} + \frac{36 0 ^{\circ} n}{3}

v = 12 0^{\circ} - \frac{36 0 ^{\circ} - 6}{3} + \frac{36 0 ^{\circ} n}{3}

v = 12 0^{\circ} - 2 + \frac{36 0 ^{\circ} n}{3}, v = 12 0^{\circ} - \frac{36 0 ^{\circ} - 6}{3} + \frac{36 0 ^{\circ} n}{3}

Solutions for the range

0 \leq v^{\circ} \leq 36 0^{\circ}

No Solution for v \in R

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sin^{2} (θ) - 7 sin (θ) = 0

- 1 = 2 cos (2 θ)

tan (x) = \frac{10}{12}

0.666 = sin (x)

What is the general solution for cos(3v)=cos(6),0<= v<= 360 ?
The general solution for cos(3v)=cos(6),0<= v<= 360 is No Solution for v\in\mathbb{R}