What is the general solution for 5sin^2(x)=0 ?

The general solution for 5sin^2(x)=0 is x=2pin,x=pi+2pin

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5sin^2(x)=0

5 sin^{2} (x) = 0

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

5 sin^{2} (x) = 0

Divide both sides by

5

5 sin^{2} (x) = 0

Divide both sides by

5

5 sin^{2} (x) = 0

Divide both sides by

5

\frac{5 sin ^{2} ( x )}{5} = \frac{0}{5}

Simplify

sin^{2} (x) = 0

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

(\frac{( 1 - sin ^{2} ( x ) )}{( 1 + 2 \cdot sin ^{2} ( x ) )}) = 0.4

cos (3 x + \frac{π}{4}) = - \frac{3}{2}

1 = cos (\frac{π}{64} \cdot t^{2})

sin (x) - \frac{1}{4} = 0

2 csc (θ) - 2 = 0

What is the general solution for 5sin^2(x)=0 ?
The general solution for 5sin^2(x)=0 is x=2pin,x=pi+2pin