What is the general solution for arccos(x)= pi/4 ?

The general solution for arccos(x)= pi/4 is x=(sqrt(2))/2

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arccos(x)= pi/4

arccos (x) = \frac{π}{4}

x = \frac{2}{2}

Solution steps

arccos (x) = \frac{π}{4}

Apply trig inverse properties

arccos (x) = \frac{π}{4}

arccos (x) = a \Rightarrow x = cos (a)

x = cos (\frac{π}{4})

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2}

x = \frac{2}{2}

x = \frac{2}{2}

sin (4 x) + cos (2 x) = 0

sin^{2} (θ) - 11 sin (θ) = 0

sin (θ) = m

sin (θ) = \frac{3}{2}, cos (θ) < 0, 0 < θ < 2 π

tan (x) = 1, 0 \leq x < 2 π

What is the general solution for arccos(x)= pi/4 ?
The general solution for arccos(x)= pi/4 is x=(sqrt(2))/2