What is the general solution for 2sin^2(x)=1,0<= x<= 2pi ?

The general solution for 2sin^2(x)=1,0<= x<= 2pi is x= pi/4 ,x=(3pi)/4 ,x=(5pi)/4 ,x=(7pi)/4

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2sin^2(x)=1,0<= x<= 2pi

2 sin^{2} (x) = 1, 0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Degrees

x = 4 5^{\circ}, x = 13 5^{\circ}, x = 22 5^{\circ}, x = 31 5^{\circ}

Solution steps

2 sin^{2} (x) = 1, 0 \leq x \leq 2 π

Solve by substitution

2 sin^{2} (x) = 1

Let:

sin (x) = u

2 u^{2} = 1

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

2 u^{2} = 1

Divide both sides by

2

2 u^{2} = 1

Divide both sides by

2

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

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

Substitute back

u = sin (x)

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

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

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π

General solutions for

sin (x) = \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}

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

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{5 π}{4}, x = \frac{7 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π

General solutions for

sin (x) = - \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}

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

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{5 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

0 = \frac{1}{5} sin (5 t)

tan^{2} (x) - 1 = 0, 0 \leq x < 2 π

6 sin^{2} (x) - 5 sin (x) + 1 = 0

cos (θ) + 1 = sin (θ)

2 sec^{2} (x) - 2 = 0

What is the general solution for 2sin^2(x)=1,0<= x<= 2pi ?
The general solution for 2sin^2(x)=1,0<= x<= 2pi is x= pi/4 ,x=(3pi)/4 ,x=(5pi)/4 ,x=(7pi)/4