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

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

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2sin^2(x)=0

2 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

2 sin^{2} (x) = 0

Divide both sides by

2

2 sin^{2} (x) = 0

Divide both sides by

2

2 sin^{2} (x) = 0

Divide both sides by

2

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

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

tan (x) = sin (x)

csc (x) = sec (x)

sec (x) = 3

- sin (x) = cos (x)

sin (3 x) + sin (x) = 0, 0 \leq x \leq 2 π

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