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

The general solution for 2sin^2(x)=sqrt(2)sin(x) is x=2pin,x=pi+2pin,x= pi/4+2pin,x=(3pi)/4+2pin

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2sin^2(x)=sqrt(2)sin(x)

Solution

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

Solution

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

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

2 u^{2} = 2 u

2 u^{2} = 2 u : u = 0, u = \frac{2}{2}

2 u^{2} = 2 u

Move

2 u

to the left side

2 u^{2} = 2 u

Subtract

2 u

from both sides

2 u^{2} - 2 u = 2 u - 2 u

Simplify

2 u^{2} - 2 u = 0

2 u^{2} - 2 u = 0

Factor

2 u^{2} - 2 u : u (2 u - 2)

2 u^{2} - 2 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= 2 uu - 2 u

Factor out common term

u

Multiply the numbers:

1 \cdot 2 = 2

= u (2 u - 2)

u (2 u - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u = 0 or 2 u - 2 = 0

Solve

2 u - 2 = 0 : u = \frac{2}{2}

2 u - 2 = 0

Move

2

to the right side

2 u - 2 = 0

Add

2

to both sides

2 u - 2 + 2 = 0 + 2

Simplify

2 u = 2

2 u = 2

Divide both sides by

2

2 u = 2

Divide both sides by

2

\frac{2 u}{2} = \frac{2}{2}

Simplify

u = \frac{2}{2}

u = \frac{2}{2}

The solutions to the quadratic equation are:

u = 0, u = \frac{2}{2}

Substitute back

u = sin (x)

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

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

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

sin (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

General solutions for

sin (x) = \frac{2}{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

Combine all the solutions

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

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tan (a) = 3

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tan (x) = \frac{6}{8}

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sin (\frac{1}{2} x) + cos (x) = 1

cos(x)tan^2(x)=cos(x)

cos (x) tan^{2} (x) = cos (x)

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)=sqrt(2)sin(x) ?
The general solution for 2sin^2(x)=sqrt(2)sin(x) is x=2pin,x=pi+2pin,x= pi/4+2pin,x=(3pi)/4+2pin