What is the general solution for 2sin^2(θ)-5sin(θ)+3=0 ?

The general solution for 2sin^2(θ)-5sin(θ)+3=0 is θ= pi/2+2pin

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Popular >

2sin^2(θ)-5sin(θ)+3=0

Solution

2 sin^{2} (θ) - 5 sin (θ) + 3 = 0

Solution

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (θ) - 5 sin (θ) + 3 = 0

Solve by substitution

2 sin^{2} (θ) - 5 sin (θ) + 3 = 0

Let:

sin (θ) = u

2 u^{2} - 5 u + 3 = 0

2 u^{2} - 5 u + 3 = 0 : u = \frac{3}{2}, u = 1

2 u^{2} - 5 u + 3 = 0

Solve with the quadratic formula

2 u^{2} - 5 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = - 5, c = 3

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 \cdot 3 = 1

(- 5)^{2} - 4 \cdot 2 \cdot 3

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 2}

u = \frac{- ( - 5 ) + 1}{2 \cdot 2} : \frac{3}{2}

\frac{- ( - 5 ) + 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 1}{2 \cdot 2}

Add the numbers:

5 + 1 = 6

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

u = \frac{- ( - 5 ) - 1}{2 \cdot 2} : 1

\frac{- ( - 5 ) - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 1}{2 \cdot 2}

Subtract the numbers:

5 - 1 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{3}{2}, u = 1

Substitute back

u = sin (θ)

sin (θ) = \frac{3}{2}, sin (θ) = 1

sin (θ) = \frac{3}{2}, sin (θ) = 1

sin (θ) = \frac{3}{2} :

No Solution

sin (θ) = \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = 1

General solutions for

sin (θ) = 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}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn

Popular Examples

2sin(x)csc(x)-csc(x)=4sin(x)-2

2 sin (x) csc (x) - csc (x) = 4 sin (x) - 2

2cos^2(x)+5cos(x)+2=0

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

sin(θ)=-0.66

sin (θ) = - 0.66

4sec^2(θ)tan(θ)=8tan(θ)

4 sec^{2} (θ) tan (θ) = 8 tan (θ)

5sec(x)csc(x)=10csc(x)

5 sec (x) csc (x) = 10 csc (x)

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(θ)-5sin(θ)+3=0 ?
The general solution for 2sin^2(θ)-5sin(θ)+3=0 is θ= pi/2+2pin