What is the general solution for 9-12sin(x)=4cos^2(x) ?

The general solution for 9-12sin(x)=4cos^2(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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9-12sin(x)=4cos^2(x)

Solution

9 - 12 sin (x) = 4 cos^{2} (x)

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

9 - 12 sin (x) = 4 cos^{2} (x)

Subtract

4 cos^{2} (x)

from both sides

9 - 12 sin (x) - 4 cos^{2} (x) = 0

Rewrite using trig identities

9 - 12 sin (x) - 4 cos^{2} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 9 - 12 sin (x) - 4 (1 - sin^{2} (x))

Simplify

9 - 12 sin (x) - 4 (1 - sin^{2} (x)) : 4 sin^{2} (x) - 12 sin (x) + 5

9 - 12 sin (x) - 4 (1 - sin^{2} (x))

Expand

- 4 (1 - sin^{2} (x)) : - 4 + 4 sin^{2} (x)

- 4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

= - 4 \cdot 1 - (- 4) sin^{2} (x)

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (x)

= 9 - 12 sin (x) - 4 + 4 sin^{2} (x)

Simplify

9 - 12 sin (x) - 4 + 4 sin^{2} (x) : 4 sin^{2} (x) - 12 sin (x) + 5

9 - 12 sin (x) - 4 + 4 sin^{2} (x)

Group like terms

= - 12 sin (x) + 4 sin^{2} (x) + 9 - 4

Add/Subtract the numbers:

9 - 4 = 5

= 4 sin^{2} (x) - 12 sin (x) + 5

= 4 sin^{2} (x) - 12 sin (x) + 5

= 4 sin^{2} (x) - 12 sin (x) + 5

5 - 12 sin (x) + 4 sin^{2} (x) = 0

Solve by substitution

5 - 12 sin (x) + 4 sin^{2} (x) = 0

Let:

sin (x) = u

5 - 12 u + 4 u^{2} = 0

5 - 12 u + 4 u^{2} = 0 : u = \frac{5}{2}, u = \frac{1}{2}

5 - 12 u + 4 u^{2} = 0

Write in the standard form

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

4 u^{2} - 12 u + 5 = 0

Solve with the quadratic formula

4 u^{2} - 12 u + 5 = 0

Quadratic Equation Formula:

For

a = 4, b = - 12, c = 5

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

(- 12)^{2} - 4 \cdot 4 \cdot 5 = 8

(- 12)^{2} - 4 \cdot 4 \cdot 5

Apply exponent rule:

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

n

is even

(- 12)^{2} = 1 2^{2}

= 1 2^{2} - 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 1 2^{2} - 80

1 2^{2} = 144

= 144 - 80

Subtract the numbers:

144 - 80 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 12 ) \pm 8}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 12 ) + 8}{2 \cdot 4}, u_{2} = \frac{- ( - 12 ) - 8}{2 \cdot 4}

u = \frac{- ( - 12 ) + 8}{2 \cdot 4} : \frac{5}{2}

\frac{- ( - 12 ) + 8}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{12 + 8}{2 \cdot 4}

Add the numbers:

12 + 8 = 20

= \frac{20}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{20}{8}

Cancel the common factor:

4

= \frac{5}{2}

u = \frac{- ( - 12 ) - 8}{2 \cdot 4} : \frac{1}{2}

\frac{- ( - 12 ) - 8}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{12 - 8}{2 \cdot 4}

Subtract the numbers:

12 - 8 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{5}{2}, u = \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = \frac{5}{2}, sin (x) = \frac{1}{2}

sin (x) = \frac{5}{2}, sin (x) = \frac{1}{2}

sin (x) = \frac{5}{2} :

No Solution

sin (x) = \frac{5}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

(1-sin(x))/(cos(x))=(cos(x))/(1-sin(x))

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

5sec(x)+7=-3

5 sec (x) + 7 = - 3

cos(2θ)+2cos(θ)=-1

cos (2 θ) + 2 cos (θ) = - 1

((tan(x))/1)^{-1}=(20)/(9.7)

(\frac{tan ( x )}{1})^{- 1} = \frac{20}{9.7}

1-tan(x)=sec(x)

1 - tan (x) = sec (x)

Frequently Asked Questions (FAQ)

What is the general solution for 9-12sin(x)=4cos^2(x) ?
The general solution for 9-12sin(x)=4cos^2(x) is x= pi/6+2pin,x=(5pi)/6+2pin