What is the general solution for 7sin^2(θ)-5sin(θ)=2 ?

The general solution for 7sin^2(θ)-5sin(θ)=2 is θ= pi/2+2pin,θ=-0.28975…+2pin,θ=pi+0.28975…+2pin

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7sin^2(θ)-5sin(θ)=2

Solution

7 sin^{2} (θ) - 5 sin (θ) = 2

Solution

θ = \frac{π}{2} + 2 πn, θ = - 0.28975 \dots + 2 πn, θ = π + 0.28975 \dots + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = - 16.60154 \dots^{\circ} + 36 0^{\circ} n, θ = 196.60154 \dots^{\circ} + 36 0^{\circ} n

Solution steps

7 sin^{2} (θ) - 5 sin (θ) = 2

Solve by substitution

7 sin^{2} (θ) - 5 sin (θ) = 2

Let:

sin (θ) = u

7 u^{2} - 5 u = 2

7 u^{2} - 5 u = 2 : u = 1, u = - \frac{2}{7}

7 u^{2} - 5 u = 2

Move

2

to the left side

7 u^{2} - 5 u = 2

Subtract

2

from both sides

7 u^{2} - 5 u - 2 = 2 - 2

Simplify

7 u^{2} - 5 u - 2 = 0

7 u^{2} - 5 u - 2 = 0

Solve with the quadratic formula

7 u^{2} - 5 u - 2 = 0

Quadratic Equation Formula:

For

a = 7, b = - 5, c = - 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 7 ( - 2 )}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 7 ( - 2 )}{2 \cdot 7}

(- 5)^{2} - 4 \cdot 7 (- 2) = 9

(- 5)^{2} - 4 \cdot 7 (- 2)

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 7 \cdot 2

Apply exponent rule:

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

n

is even

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

= 5^{2} + 4 \cdot 7 \cdot 2

Multiply the numbers:

4 \cdot 7 \cdot 2 = 56

= 5^{2} + 56

5^{2} = 25

= 25 + 56

Add the numbers:

25 + 56 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

u_{1, 2} = \frac{- ( - 5 ) \pm 9}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 9}{2 \cdot 7}, u_{2} = \frac{- ( - 5 ) - 9}{2 \cdot 7}

u = \frac{- ( - 5 ) + 9}{2 \cdot 7} : 1

\frac{- ( - 5 ) + 9}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{5 + 9}{2 \cdot 7}

Add the numbers:

5 + 9 = 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{- ( - 5 ) - 9}{2 \cdot 7} : - \frac{2}{7}

\frac{- ( - 5 ) - 9}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{5 - 9}{2 \cdot 7}

Subtract the numbers:

5 - 9 = - 4

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

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 4}{14}

Apply the fraction rule:

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

= - \frac{4}{14}

Cancel the common factor:

2

= - \frac{2}{7}

The solutions to the quadratic equation are:

u = 1, u = - \frac{2}{7}

Substitute back

u = sin (θ)

sin (θ) = 1, sin (θ) = - \frac{2}{7}

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

sin (θ) = - \frac{2}{7}

Apply trig inverse properties

sin (θ) = - \frac{2}{7}

General solutions for

sin (θ) = - \frac{2}{7}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

θ = \frac{π}{2} + 2 πn, θ = - 0.28975 \dots + 2 πn, θ = π + 0.28975 \dots + 2 πn

Popular Examples

sec(2x)=-(2/(sqrt(3)))

sec (2 x) = - (\frac{2}{3})

(e^{-ln(-(sin(θ))/(cos(θ)))})/2*sin(θ)=0

\frac{e ^{- l n (- \frac{s i n ( θ )}{c o s ( θ )})}}{2} \cdot sin (θ) = 0

7sin(2x)+12sin(x)=0

7 sin (2 x) + 12 sin (x) = 0

solvefor x,cos(x)=cos(2x)solve for

x, cos (x) = cos (2 x)

cos(x)=0.50074242

cos (x) = 0.50074242

Frequently Asked Questions (FAQ)

What is the general solution for 7sin^2(θ)-5sin(θ)=2 ?
The general solution for 7sin^2(θ)-5sin(θ)=2 is θ= pi/2+2pin,θ=-0.28975…+2pin,θ=pi+0.28975…+2pin