What is the general solution for 2sin^2(θ)-7sin(θ)-4=0 ?

The general solution for 2sin^2(θ)-7sin(θ)-4=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(θ)-7sin(θ)-4=0

Solution

2 sin^{2} (θ) - 7 sin (θ) - 4 = 0

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (θ) - 7 sin (θ) - 4 = 0

Solve by substitution

2 sin^{2} (θ) - 7 sin (θ) - 4 = 0

Let:

sin (θ) = u

2 u^{2} - 7 u - 4 = 0

2 u^{2} - 7 u - 4 = 0 : u = 4, u = - \frac{1}{2}

2 u^{2} - 7 u - 4 = 0

Solve with the quadratic formula

2 u^{2} - 7 u - 4 = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = - 4

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 (- 4) = 9

(- 7)^{2} - 4 \cdot 2 (- 4)

Apply rule

- (- a) = a

= (- 7)^{2} + 4 \cdot 2 \cdot 4

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 2 \cdot 4

Multiply the numbers:

4 \cdot 2 \cdot 4 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

Add the numbers:

49 + 32 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 7 ) \pm 9}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 9}{2 \cdot 2}, u_{2} = \frac{- ( - 7 ) - 9}{2 \cdot 2}

u = \frac{- ( - 7 ) + 9}{2 \cdot 2} : 4

\frac{- ( - 7 ) + 9}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 + 9}{2 \cdot 2}

Add the numbers:

7 + 9 = 16

= \frac{16}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{16}{4}

Divide the numbers:

\frac{16}{4} = 4

= 4

u = \frac{- ( - 7 ) - 9}{2 \cdot 2} : - \frac{1}{2}

\frac{- ( - 7 ) - 9}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{7 - 9}{2 \cdot 2}

Subtract the numbers:

7 - 9 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = 4, u = - \frac{1}{2}

Substitute back

u = sin (θ)

sin (θ) = 4, sin (θ) = - \frac{1}{2}

sin (θ) = 4, sin (θ) = - \frac{1}{2}

sin (θ) = 4 :

No Solution

sin (θ) = 4

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \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}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Popular Examples

2tan(x)-2sqrt(3)=0

2 tan (x) - 23 = 0

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

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

2sin^2(x/4)-3cos(x/4)=0

2 sin^{2} (\frac{x}{4}) - 3 cos (\frac{x}{4}) = 0

cos^2(A)+sin^2(A)= 1/(sin^2(A))

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

cos(4θ)=1

cos (4 θ) = 1

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(θ)-7sin(θ)-4=0 ?
The general solution for 2sin^2(θ)-7sin(θ)-4=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin