What is the general solution for cos(x^2)=1 ?

The general solution for cos(x^2)=1 is x=sqrt(2)sqrt(pin),x=-sqrt(2)sqrt(pin)

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cos(x^2)=1

cos (x^{2}) = 1

x = 2 πn, x = - 2 πn

Degrees

x = 0^{\circ} + 143.61922 \dots^{\circ} n, x = 0^{\circ} - 143.61922 \dots^{\circ} n

Solution steps

cos (x^{2}) = 1

General solutions for

cos (x^{2}) = 1

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}

x^{2} = 0 + 2 πn

x^{2} = 0 + 2 πn

Solve

x^{2} = 0 + 2 πn : x = 2 πn, x = - 2 πn

x^{2} = 0 + 2 πn

0 + 2 πn = 2 πn

x^{2} = 2 πn

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 2 πn, x = - 2 πn

Simplify

2 πn : 2 πn

2 πn

Apply radical rule:

assuming

a \geq 0, b \geq 0

= 2 πn

Simplify

- 2 πn : - 2 πn

- 2 πn

Apply radical rule:

assuming

a \geq 0, b \geq 0

= - 2 πn

x = 2 πn, x = - 2 πn

x = 2 πn, x = - 2 πn

2 sin^{2} (θ) + cos (θ) - 2 = 0

sin^{3} (x) - 0.25 sin (x) = 0

sin^{2} (x) cos (x) = 0

sin (x) - cos (x) = \frac{3}{2}

cos (x) = \frac{1 - cos ( x )}{2}

What is the general solution for cos(x^2)=1 ?
The general solution for cos(x^2)=1 is x=sqrt(2)sqrt(pin),x=-sqrt(2)sqrt(pin)