What is the general solution for 4sin^2(x)+1=4sin(x) ?

The general solution for 4sin^2(x)+1=4sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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4sin^2(x)+1=4sin(x)

Solution

4 sin^{2} (x) + 1 = 4 sin (x)

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{2} (x) + 1 = 4 sin (x)

Solve by substitution

4 sin^{2} (x) + 1 = 4 sin (x)

Let:

sin (x) = u

4 u^{2} + 1 = 4 u

4 u^{2} + 1 = 4 u : u = \frac{1}{2}

4 u^{2} + 1 = 4 u

Move

4 u

to the left side

4 u^{2} + 1 = 4 u

Subtract

4 u

from both sides

4 u^{2} + 1 - 4 u = 4 u - 4 u

Simplify

4 u^{2} + 1 - 4 u = 0

4 u^{2} + 1 - 4 u = 0

Write in the standard form

a x^{2} + b x + c = 0

4 u^{2} - 4 u + 1 = 0

Solve with the quadratic formula

4 u^{2} - 4 u + 1 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{1}{2}

The solution to the quadratic equation is:

u = \frac{1}{2}

Substitute back

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

-0.6=sin(30t)

- 0.6 = sin (30 t)

1=tan(pi/4+x)

1 = tan (\frac{π}{4} + x)

tan(x)=(6.1)/4

tan (x) = \frac{6.1}{4}

4cos(θ)=4-4cos(θ)

4 cos (θ) = 4 - 4 cos (θ)

sin^2(x)+6sin(x)+5=0

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

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)+1=4sin(x) ?
The general solution for 4sin^2(x)+1=4sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin