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Popular >

3sin^2(x)+6sin(x)-11=7sin(x)-9

Solution

3 sin^{2} (x) + 6 sin (x) - 11 = 7 sin (x) - 9

Solution

x = \frac{π}{2} + 2 πn, x = - 0.72972 \dots + 2 πn, x = π + 0.72972 \dots + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = - 41.81031 \dots^{\circ} + 36 0^{\circ} n, x = 221.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 sin^{2} (x) + 6 sin (x) - 11 = 7 sin (x) - 9

Solve by substitution

3 sin^{2} (x) + 6 sin (x) - 11 = 7 sin (x) - 9

Let:

sin (x) = u

3 u^{2} + 6 u - 11 = 7 u - 9

3 u^{2} + 6 u - 11 = 7 u - 9 : u = 1, u = - \frac{2}{3}

3 u^{2} + 6 u - 11 = 7 u - 9

Move

9

to the left side

3 u^{2} + 6 u - 11 = 7 u - 9

Add

9

to both sides

3 u^{2} + 6 u - 11 + 9 = 7 u - 9 + 9

Simplify

3 u^{2} + 6 u - 2 = 7 u

3 u^{2} + 6 u - 2 = 7 u

Move

7 u

to the left side

3 u^{2} + 6 u - 2 = 7 u

Subtract

7 u

from both sides

3 u^{2} + 6 u - 2 - 7 u = 7 u - 7 u

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 3, b = - 1, c = - 2

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

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

(- 1)^{2} - 4 \cdot 3 (- 2) = 5

(- 1)^{2} - 4 \cdot 3 (- 2)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 3 \cdot 2

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 3 \cdot 2 = 24

4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

= \frac{1 + 5}{2 \cdot 3}

Add the numbers:

1 + 5 = 6

= \frac{6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

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

\frac{- ( - 1 ) - 5}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{1 - 5}{2 \cdot 3}

Subtract the numbers:

1 - 5 = - 4

= \frac{- 4}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 4}{6}

Apply the fraction rule:

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

= - \frac{4}{6}

Cancel the common factor:

2

= - \frac{2}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = - 0.72972 \dots + 2 πn, x = π + 0.72972 \dots + 2 πn

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