What is the general solution for 3sin^2(θ)-7sin(θ)+2=0 ?

The general solution for 3sin^2(θ)-7sin(θ)+2=0 is θ=0.33983…+2pin,θ=pi-0.33983…+2pin

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Popular >

3sin^2(θ)-7sin(θ)+2=0

Solution

3 sin^{2} (θ) - 7 sin (θ) + 2 = 0

Solution

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn

Degrees

θ = 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 160.52877 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin^{2} (θ) - 7 sin (θ) + 2 = 0

Solve by substitution

3 sin^{2} (θ) - 7 sin (θ) + 2 = 0

Let:

sin (θ) = u

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

3 u^{2} - 7 u + 2 = 0 : u = 2, u = \frac{1}{3}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 3, b = - 7, c = 2

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

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

(- 7)^{2} - 4 \cdot 3 \cdot 2 = 5

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 3}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 3}

u = \frac{- ( - 7 ) + 5}{2 \cdot 3} : 2

\frac{- ( - 7 ) + 5}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{7 + 5}{2 \cdot 3}

Add the numbers:

7 + 5 = 12

= \frac{12}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{12}{6}

Divide the numbers:

\frac{12}{6} = 2

= 2

u = \frac{- ( - 7 ) - 5}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 7 ) - 5}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{7 - 5}{2 \cdot 3}

Subtract the numbers:

7 - 5 = 2

= \frac{2}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{2}{6}

Cancel the common factor:

2

= \frac{1}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = 2, sin (θ) = \frac{1}{3}

sin (θ) = 2, sin (θ) = \frac{1}{3}

sin (θ) = 2 :

No Solution

sin (θ) = 2

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{1}{3} : θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

sin (θ) = \frac{1}{3}

Apply trig inverse properties

sin (θ) = \frac{1}{3}

General solutions for

sin (θ) = \frac{1}{3}

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

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

Show solutions in decimal form

θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn

Popular Examples

9sin^2(θ)+14cos(θ)-12=5cos(θ)-1

9 sin^{2} (θ) + 14 cos (θ) - 12 = 5 cos (θ) - 1

4sec^2(x)-2=0

4 sec^{2} (x) - 2 = 0

cos(x)= 1/(sqrt(3))

cos (x) = \frac{1}{3}

sin(x)cos(x)=(sqrt(3))/4 ,0<= x<= 2pi

sin (x) cos (x) = \frac{3}{4}, 0 \leq x \leq 2 π

2cos^2(x)+sin(x)-1=0,0<= x<2pi

2 cos^{2} (x) + sin (x) - 1 = 0, 0 \leq x < 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(θ)-7sin(θ)+2=0 ?
The general solution for 3sin^2(θ)-7sin(θ)+2=0 is θ=0.33983…+2pin,θ=pi-0.33983…+2pin