What is the general solution for 2sin^2(u)=1+,u ?

The general solution for 2sin^2(u)=1+,u is No Solution for u\in\mathbb{R}

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2sin^2(u)=1+,u

2 sin^{2} (u) = 1 +, u

No Solution for u \in R

Solution steps

2 sin^{2} (u) = 1 +, u

Solve by substitution

2 sin^{2} (u) = 1

Let:

sin (u) = v

2 v^{2} = 1

2 v^{2} = 1 : v = \frac{1}{2}, v = - \frac{1}{2}

2 v^{2} = 1

Divide both sides by

2

2 v^{2} = 1

Divide both sides by

2

\frac{2 v ^{2}}{2} = \frac{1}{2}

Simplify

v^{2} = \frac{1}{2}

v^{2} = \frac{1}{2}

For

x^{2} = f (a)

the solutions are

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

v = \frac{1}{2}, v = - \frac{1}{2}

Substitute back

v = sin (u)

sin (u) = \frac{1}{2}, sin (u) = - \frac{1}{2}

sin (u) = \frac{1}{2}, sin (u) = - \frac{1}{2}

sin (u) = \frac{1}{2}, u :

No Solution

sin (u) = \frac{1}{2}, u

General solutions for

sin (u) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

u = \frac{π}{4} + 2 πn, u = \frac{3 π}{4} + 2 πn

u = \frac{π}{4} + 2 πn, u = \frac{3 π}{4} + 2 πn

Solutions for the range

u

No Solution

sin (u) = - \frac{1}{2}, u :

No Solution

sin (u) = - \frac{1}{2}, u

General solutions for

sin (u) = - \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

u = \frac{5 π}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn

u = \frac{5 π}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn

Solutions for the range

u

No Solution

Combine all the solutions

No Solution for u \in R

8.5 = 12 sin (a)

cos (x) = \frac{1}{15}

cot (θ) = - 3, 0 \leq θ < 2 π

\frac{sin ( 9 0 ^{\circ} )}{458.912} = \frac{sin ( x )}{450}

0 = \frac{2 π ^{2} cos ( \frac{π x}{10} )}{25}

What is the general solution for 2sin^2(u)=1+,u ?
The general solution for 2sin^2(u)=1+,u is No Solution for u\in\mathbb{R}