What is the general solution for sin^2(a)-4sin(a)+3=0 ?

The general solution for sin^2(a)-4sin(a)+3=0 is a= pi/2+2pin

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Popular >

sin^2(a)-4sin(a)+3=0

Solution

sin^{2} (a) - 4 sin (a) + 3 = 0

Solution

a = \frac{π}{2} + 2 πn

Degrees

a = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (a) - 4 sin (a) + 3 = 0

Solve by substitution

sin^{2} (a) - 4 sin (a) + 3 = 0

Let:

sin (a) = u

u^{2} - 4 u + 3 = 0

u^{2} - 4 u + 3 = 0 : u = 3, u = 1

u^{2} - 4 u + 3 = 0

Solve with the quadratic formula

u^{2} - 4 u + 3 = 0

Quadratic Equation Formula:

For

a = 1, b = - 4, c = 3

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 3 = 2

(- 4)^{2} - 4 \cdot 1 \cdot 3

Apply exponent rule:

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

n

is even

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

= 4^{2} - 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

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

u = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

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

Apply rule

- (- a) = a

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

Add the numbers:

4 + 2 = 6

= \frac{6}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

4 - 2 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 3, u = 1

Substitute back

u = sin (a)

sin (a) = 3, sin (a) = 1

sin (a) = 3, sin (a) = 1

sin (a) = 3 :

No Solution

sin (a) = 3

- 1 \leq sin (x) \leq 1

No Solution

sin (a) = 1 : a = \frac{π}{2} + 2 πn

sin (a) = 1

General solutions for

sin (a) = 1

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}

a = \frac{π}{2} + 2 πn

a = \frac{π}{2} + 2 πn

Combine all the solutions

a = \frac{π}{2} + 2 πn

Popular Examples

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tan(a)+sec(a)= 3/2

tan (a) + sec (a) = \frac{3}{2}

3cos^2(x)+4cos^4(x)-5=0

3 cos^{2} (x) + 4 cos^{4} (x) - 5 = 0

cos(2x+3)=0

cos (2 x + 3) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(a)-4sin(a)+3=0 ?
The general solution for sin^2(a)-4sin(a)+3=0 is a= pi/2+2pin