What is the general solution for solvefor a,90=180-2*arccos((2a)/c) ?

The general solution for solvefor a,90=180-2*arccos((2a)/c) is a= c/(2sqrt(2))

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solvefor a,90=180-2*arccos((2a)/c)

solve for

a, 9 0^{\circ} = 18 0^{\circ} - 2 \cdot arccos (\frac{2 a}{c})

a = \frac{c}{2 2}

Solution steps

9 0^{\circ} = 18 0^{\circ} - 2 arccos (\frac{2 a}{c})

Switch sides

π - 2 arccos (\frac{2 a}{c}) = \frac{π}{2}

Solve by substitution

arccos (\frac{2 a}{c}) = \frac{π}{4}

Subtract

9 0^{\circ}

from both sides

π - 2 arccos (\frac{2 a}{c}) - \frac{π}{2} = 0

Simplify

18 0^{\circ} - 2 arccos (\frac{2 a}{c}) - 9 0^{\circ} : \frac{18 0 ^{\circ} - 4 arccos ( \frac{2 a}{c} )}{2}

\frac{π - 4 arccos ( \frac{2 a}{c} )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

π - 4 arccos (\frac{2 a}{c}) = 0

Solve by substitution

arccos (\frac{2 a}{c}) = \frac{π}{4}

Let:

u = \frac{2 a}{c}

18 0^{\circ} - 4 arccos (u) = 0

Move

18 0^{\circ}

to the right side

- 4 arccos (u) = - 18 0^{\circ}

Divide both sides by

- 4

arccos (u) = 4 5^{\circ}

Apply trig inverse properties

u = \frac{2}{2}

Substitute back

u = \frac{2 a}{c}

\frac{2 a}{c} = \frac{2}{2} : a = \frac{c}{2 2}; c \neq = 0

a = \frac{c}{2 2}

cos (x) = \frac{π}{4}

cot (x) = \frac{3}{7}, cos (x) > 0

4 tanh (x) - \frac{1}{cosh ( x )} = 1

2 sin (θ) cos (θ) = sin (θ)

53 cot (x) + 5 = 0

What is the general solution for solvefor a,90=180-2*arccos((2a)/c) ?
The general solution for solvefor a,90=180-2*arccos((2a)/c) is a= c/(2sqrt(2))