What is the general solution for cos(2B)= 21/35 ,cos(B) ?

The general solution for cos(2B)= 21/35 ,cos(B) is No Solution for B\in\mathbb{R}

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cos(2B)= 21/35 ,cos(B)

cos (2 B) = \frac{21}{35}, cos (B)

No Solution for B \in R

Solution steps

cos (2 B) = \frac{21}{35}, cos (B)

Cancel the common factor:

7

cos (2 B) = \frac{3}{5}

Apply trig inverse properties

cos (2 B) = \frac{3}{5}

General solutions for

cos (2 B) = \frac{3}{5}

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

2 B = arccos (\frac{3}{5}) + 2 πn, 2 B = 2 π - arccos (\frac{3}{5}) + 2 πn

2 B = arccos (\frac{3}{5}) + 2 πn, 2 B = 2 π - arccos (\frac{3}{5}) + 2 πn

Solve

2 B = arccos (\frac{3}{5}) + 2 πn : B = \frac{arccos ( \frac{3}{5} )}{2} + πn

2 B = arccos (\frac{3}{5}) + 2 πn

Divide both sides by

2

2 B = arccos (\frac{3}{5}) + 2 πn

Divide both sides by

2

\frac{2 B}{2} = \frac{arccos ( \frac{3}{5} )}{2} + \frac{2 πn}{2}

Simplify

B = \frac{arccos ( \frac{3}{5} )}{2} + πn

B = \frac{arccos ( \frac{3}{5} )}{2} + πn

Solve

2 B = 2 π - arccos (\frac{3}{5}) + 2 πn : B = π - \frac{arccos ( \frac{3}{5} )}{2} + πn

2 B = 2 π - arccos (\frac{3}{5}) + 2 πn

Divide both sides by

2

2 B = 2 π - arccos (\frac{3}{5}) + 2 πn

Divide both sides by

2

\frac{2 B}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{3}{5} )}{2} + \frac{2 πn}{2}

Simplify

B = π - \frac{arccos ( \frac{3}{5} )}{2} + πn

B = π - \frac{arccos ( \frac{3}{5} )}{2} + πn

B = \frac{arccos ( \frac{3}{5} )}{2} + πn, B = π - \frac{arccos ( \frac{3}{5} )}{2} + πn

Solutions for the range

cos (B)

No Solution for B \in R

2 cos^{4} (x) + 3 sin^{2} (x) - 2 = 0

cos^{2} (x) + 4 sin^{2} (\frac{x}{2}) = 1

4 cos (x) = sin^{2} (x) + 1

cos^{2} (x) + sin^{2} (x) - 2 sin (x) = 0, 0 \leq x \leq 2 π

cot (A) = \frac{24}{7}

What is the general solution for cos(2B)= 21/35 ,cos(B) ?
The general solution for cos(2B)= 21/35 ,cos(B) is No Solution for B\in\mathbb{R}