What is the general solution for 3cos^2(x)+4sin(x)-4=0 ?

The general solution for 3cos^2(x)+4sin(x)-4=0 is x=0.33983…+2pin,x=pi-0.33983…+2pin,x= pi/2+2pin

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3cos^2(x)+4sin(x)-4=0

Solution

3 cos^{2} (x) + 4 sin (x) - 4 = 0

Solution

x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn, x = \frac{π}{2} + 2 πn

Degrees

x = 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 160.52877 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (x) + 4 sin (x) - 4 = 0

Rewrite using trig identities

- 4 + 3 cos^{2} (x) + 4 sin (x)

Use the Pythagorean identity:

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

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

= - 4 + 3 (1 - sin^{2} (x)) + 4 sin (x)

Simplify

- 4 + 3 (1 - sin^{2} (x)) + 4 sin (x) : 4 sin (x) - 3 sin^{2} (x) - 1

- 4 + 3 (1 - sin^{2} (x)) + 4 sin (x)

Expand

3 (1 - sin^{2} (x)) : 3 - 3 sin^{2} (x)

3 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

= 3 \cdot 1 - 3 sin^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (x)

= - 4 + 3 - 3 sin^{2} (x) + 4 sin (x)

Add/Subtract the numbers:

- 4 + 3 = - 1

= 4 sin (x) - 3 sin^{2} (x) - 1

= 4 sin (x) - 3 sin^{2} (x) - 1

- 1 - 3 sin^{2} (x) + 4 sin (x) = 0

Solve by substitution

- 1 - 3 sin^{2} (x) + 4 sin (x) = 0

Let:

sin (x) = u

- 1 - 3 u^{2} + 4 u = 0

- 1 - 3 u^{2} + 4 u = 0 : u = \frac{1}{3}, u = 1

- 1 - 3 u^{2} + 4 u = 0

Write in the standard form

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

- 3 u^{2} + 4 u - 1 = 0

Solve with the quadratic formula

- 3 u^{2} + 4 u - 1 = 0

Quadratic Equation Formula:

For

a = - 3, b = 4, c = - 1

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

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

4^{2} - 4 (- 3) (- 1) = 2

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

Apply rule

- (- a) = a

= 4^{2} - 4 \cdot 3 \cdot 1

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

u_{1} = \frac{- 4 + 2}{2 ( - 3 )}, u_{2} = \frac{- 4 - 2}{2 ( - 3 )}

u = \frac{- 4 + 2}{2 ( - 3 )} : \frac{1}{3}

\frac{- 4 + 2}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 4 + 2 = - 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}

u = \frac{- 4 - 2}{2 ( - 3 )} : 1

\frac{- 4 - 2}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 4 - 2 = - 6

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

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{- 6}

Apply the fraction rule:

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

= \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn, x = \frac{π}{2} + 2 πn

Popular Examples

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sin (x) = \frac{7}{15}

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3 sec (x) = 9

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

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)+4sin(x)-4=0 ?
The general solution for 3cos^2(x)+4sin(x)-4=0 is x=0.33983…+2pin,x=pi-0.33983…+2pin,x= pi/2+2pin