What is the general solution for 7sin^2(θ)-4sin(θ)=3 ?

The general solution for 7sin^2(θ)-4sin(θ)=3 is θ= pi/2+2pin,θ=-0.44291…+2pin,θ=pi+0.44291…+2pin

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7sin^2(θ)-4sin(θ)=3

Solution

7 sin^{2} (θ) - 4 sin (θ) = 3

Solution

θ = \frac{π}{2} + 2 πn, θ = - 0.44291 \dots + 2 πn, θ = π + 0.44291 \dots + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = - 25.37693 \dots^{\circ} + 36 0^{\circ} n, θ = 205.37693 \dots^{\circ} + 36 0^{\circ} n

Solution steps

7 sin^{2} (θ) - 4 sin (θ) = 3

Solve by substitution

7 sin^{2} (θ) - 4 sin (θ) = 3

Let:

sin (θ) = u

7 u^{2} - 4 u = 3

7 u^{2} - 4 u = 3 : u = 1, u = - \frac{3}{7}

7 u^{2} - 4 u = 3

Move

3

to the left side

7 u^{2} - 4 u = 3

Subtract

3

from both sides

7 u^{2} - 4 u - 3 = 3 - 3

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 7, b = - 4, c = - 3

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 7 ( - 3 )}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 7 ( - 3 )}{2 \cdot 7}

(- 4)^{2} - 4 \cdot 7 (- 3) = 10

(- 4)^{2} - 4 \cdot 7 (- 3)

Apply rule

- (- a) = a

= (- 4)^{2} + 4 \cdot 7 \cdot 3

Apply exponent rule:

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

n

is even

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

= 4^{2} + 4 \cdot 7 \cdot 3

Multiply the numbers:

4 \cdot 7 \cdot 3 = 84

= 4^{2} + 84

4^{2} = 16

= 16 + 84

Add the numbers:

16 + 84 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- ( - 4 ) \pm 10}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 10}{2 \cdot 7}, u_{2} = \frac{- ( - 4 ) - 10}{2 \cdot 7}

u = \frac{- ( - 4 ) + 10}{2 \cdot 7} : 1

\frac{- ( - 4 ) + 10}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{4 + 10}{2 \cdot 7}

Add the numbers:

4 + 10 = 14

= \frac{14}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{14}{14}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 4 ) - 10}{2 \cdot 7} : - \frac{3}{7}

\frac{- ( - 4 ) - 10}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{4 - 10}{2 \cdot 7}

Subtract the numbers:

4 - 10 = - 6

= \frac{- 6}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 6}{14}

Apply the fraction rule:

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

= - \frac{6}{14}

Cancel the common factor:

2

= - \frac{3}{7}

The solutions to the quadratic equation are:

u = 1, u = - \frac{3}{7}

Substitute back

u = sin (θ)

sin (θ) = 1, sin (θ) = - \frac{3}{7}

sin (θ) = 1, sin (θ) = - \frac{3}{7}

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

sin (θ) = - \frac{3}{7} : θ = arcsin (- \frac{3}{7}) + 2 πn, θ = π + arcsin (\frac{3}{7}) + 2 πn

sin (θ) = - \frac{3}{7}

Apply trig inverse properties

sin (θ) = - \frac{3}{7}

General solutions for

sin (θ) = - \frac{3}{7}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (- \frac{3}{7}) + 2 πn, θ = π + arcsin (\frac{3}{7}) + 2 πn

θ = arcsin (- \frac{3}{7}) + 2 πn, θ = π + arcsin (\frac{3}{7}) + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = arcsin (- \frac{3}{7}) + 2 πn, θ = π + arcsin (\frac{3}{7}) + 2 πn

Show solutions in decimal form

θ = \frac{π}{2} + 2 πn, θ = - 0.44291 \dots + 2 πn, θ = π + 0.44291 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 7sin^2(θ)-4sin(θ)=3 ?
The general solution for 7sin^2(θ)-4sin(θ)=3 is θ= pi/2+2pin,θ=-0.44291…+2pin,θ=pi+0.44291…+2pin