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

The general solution for 3sin^2(x)=5sin(x)+2 is x=-0.33983…+2pin,x=pi+0.33983…+2pin

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3sin^2(x)=5sin(x)+2

Solution

3 sin^{2} (x) = 5 sin (x) + 2

Solution

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

Degrees

x = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 199.47122 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin^{2} (x) = 5 sin (x) + 2

Solve by substitution

3 sin^{2} (x) = 5 sin (x) + 2

Let:

sin (x) = u

3 u^{2} = 5 u + 2

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

3 u^{2} = 5 u + 2

Move

2

to the left side

3 u^{2} = 5 u + 2

Subtract

2

from both sides

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

Simplify

3 u^{2} - 2 = 5 u

3 u^{2} - 2 = 5 u

Move

5 u

to the left side

3 u^{2} - 2 = 5 u

Subtract

5 u

from both sides

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

Simplify

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

5 - 7 = - 2

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{6}

Apply the fraction rule:

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

= - \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 (x)

sin (x) = 2, sin (x) = - \frac{1}{3}

sin (x) = 2, sin (x) = - \frac{1}{3}

sin (x) = 2 :

No Solution

sin (x) = 2

- 1 \leq sin (x) \leq 1

No Solution

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

sin (x) = - \frac{1}{3}

Apply trig inverse properties

sin (x) = - \frac{1}{3}

General solutions for

sin (x) = - \frac{1}{3}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

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

Popular Examples

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

sin(x)=(sqrt(7))/4

sin (x) = \frac{7}{4}

2tan^2(x)+9tan(x)+3=0

2 tan^{2} (x) + 9 tan (x) + 3 = 0

sin^2(x)=((2m-4)/5)

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1.25cos^2(θ)+0.694cos(θ)-0.519=0

1.25 cos^{2} (θ) + 0.694 cos (θ) - 0.519 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(x)=5sin(x)+2 ?
The general solution for 3sin^2(x)=5sin(x)+2 is x=-0.33983…+2pin,x=pi+0.33983…+2pin