What is the general solution for sin^2(x)-4sin(x)=0 ?

The general solution for sin^2(x)-4sin(x)=0 is x=2pin,x=pi+2pin

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Popular >

sin^2(x)-4sin(x)=0

Solution

sin^{2} (x) - 4 sin (x) = 0

Solution

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

sin^{2} (x) - 4 sin (x) = 0

Solve by substitution

sin^{2} (x) - 4 sin (x) = 0

Let:

sin (x) = u

u^{2} - 4 u = 0

u^{2} - 4 u = 0 : u = 4, u = 0

u^{2} - 4 u = 0

Solve with the quadratic formula

u^{2} - 4 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 4, c = 0

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 0 = 4

(- 4)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 4^{2} - 0

4^{2} - 0 = 4^{2}

= 4^{2}

Apply radical rule:

assuming

a \geq 0

= 4

u_{1, 2} = \frac{- ( - 4 ) \pm 4}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 4}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 4}{2 \cdot 1}

u = \frac{- ( - 4 ) + 4}{2 \cdot 1} : 4

\frac{- ( - 4 ) + 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 + 4}{2 \cdot 1}

Add the numbers:

4 + 4 = 8

= \frac{8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4

u = \frac{- ( - 4 ) - 4}{2 \cdot 1} : 0

\frac{- ( - 4 ) - 4}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 - 4}{2 \cdot 1}

Subtract the numbers:

4 - 4 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 4, u = 0

Substitute back

u = sin (x)

sin (x) = 4, sin (x) = 0

sin (x) = 4, sin (x) = 0

sin (x) = 4 :

No Solution

sin (x) = 4

- 1 \leq sin (x) \leq 1

No Solution

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

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

sin^2(x)-cos(x)-1=0

sin^{2} (x) - cos (x) - 1 = 0

sqrt(3)tan(2θ)+1=0

3 tan (2 θ) + 1 = 0

sin^2(x)-cos(x)=0

sin^{2} (x) - cos (x) = 0

9sin(x)=cos(x)-7

9 sin (x) = cos (x) - 7

tan(θ)= 1/(-sqrt(3))

tan (θ) = \frac{1}{- 3}

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-4sin(x)=0 ?
The general solution for sin^2(x)-4sin(x)=0 is x=2pin,x=pi+2pin