What is the general solution for 3cos^2(x)+5sin(x)-4=0 ?

The general solution for 3cos^2(x)+5sin(x)-4=0 is x=0.23455…+2pin,x=pi-0.23455…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3cos^2(x)+5sin(x)-4=0

Solution

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

Solution

x = 0.23455 \dots + 2 πn, x = π - 0.23455 \dots + 2 πn

Degrees

x = 13.43888 \dots^{\circ} + 36 0^{\circ} n, x = 166.56111 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

- 4 + 3 cos^{2} (x) + 5 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)) + 5 sin (x)

Simplify

- 4 + 3 (1 - sin^{2} (x)) + 5 sin (x) : 5 sin (x) - 3 sin^{2} (x) - 1

- 4 + 3 (1 - sin^{2} (x)) + 5 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) + 5 sin (x)

Add/Subtract the numbers:

- 4 + 3 = - 1

= 5 sin (x) - 3 sin^{2} (x) - 1

= 5 sin (x) - 3 sin^{2} (x) - 1

- 1 - 3 sin^{2} (x) + 5 sin (x) = 0

Solve by substitution

- 1 - 3 sin^{2} (x) + 5 sin (x) = 0

Let:

sin (x) = u

- 1 - 3 u^{2} + 5 u = 0

- 1 - 3 u^{2} + 5 u = 0 : u = \frac{5 - 13}{6}, u = \frac{5 + 13}{6}

- 1 - 3 u^{2} + 5 u = 0

Write in the standard form

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

- 3 u^{2} + 5 u - 1 = 0

Solve with the quadratic formula

- 3 u^{2} + 5 u - 1 = 0

Quadratic Equation Formula:

For

a = - 3, b = 5, c = - 1

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

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

5^{2} - 4 (- 3) (- 1) = 13

5^{2} - 4 (- 3) (- 1)

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 3 \cdot 1 = 12

= 5^{2} - 12

5^{2} = 25

= 25 - 12

Subtract the numbers:

25 - 12 = 13

= 13

u_{1, 2} = \frac{- 5 \pm 13}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- 5 + 13}{2 ( - 3 )}, u_{2} = \frac{- 5 - 13}{2 ( - 3 )}

u = \frac{- 5 + 13}{2 ( - 3 )} : \frac{5 - 13}{6}

\frac{- 5 + 13}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 13}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 5 + 13}{- 6}

Apply the fraction rule:

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

- 5 + 13 = - (5 - 13)

= \frac{5 - 13}{6}

u = \frac{- 5 - 13}{2 ( - 3 )} : \frac{5 + 13}{6}

\frac{- 5 - 13}{2 ( - 3 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 13}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 5 - 13}{- 6}

Apply the fraction rule:

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

- 5 - 13 = - (5 + 13)

= \frac{5 + 13}{6}

The solutions to the quadratic equation are:

u = \frac{5 - 13}{6}, u = \frac{5 + 13}{6}

Substitute back

u = sin (x)

sin (x) = \frac{5 - 13}{6}, sin (x) = \frac{5 + 13}{6}

sin (x) = \frac{5 - 13}{6}, sin (x) = \frac{5 + 13}{6}

sin (x) = \frac{5 - 13}{6} : x = arcsin (\frac{5 - 13}{6}) + 2 πn, x = π - arcsin (\frac{5 - 13}{6}) + 2 πn

sin (x) = \frac{5 - 13}{6}

Apply trig inverse properties

sin (x) = \frac{5 - 13}{6}

General solutions for

sin (x) = \frac{5 - 13}{6}

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

x = arcsin (\frac{5 - 13}{6}) + 2 πn, x = π - arcsin (\frac{5 - 13}{6}) + 2 πn

x = arcsin (\frac{5 - 13}{6}) + 2 πn, x = π - arcsin (\frac{5 - 13}{6}) + 2 πn

sin (x) = \frac{5 + 13}{6} :

No Solution

sin (x) = \frac{5 + 13}{6}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (\frac{5 - 13}{6}) + 2 πn, x = π - arcsin (\frac{5 - 13}{6}) + 2 πn

Show solutions in decimal form

x = 0.23455 \dots + 2 πn, x = π - 0.23455 \dots + 2 πn

Popular Examples

solvefor x,d^2+3d+2y=4cos^2(x)solve for

x, d^{2} + 3 d + 2 y = 4 cos^{2} (x)

(1+tan^2(x))cos^2(x)=sec(x)

(1 + tan^{2} (x)) cos^{2} (x) = sec (x)

(cos^2(x)-1)/(sin^2(x))=-tan(x)

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

cos^6(x)=-cos^2(x)

cos^{6} (x) = - cos^{2} (x)

2sin^3(x)-5sin^2(x)+2sin(x)=0

2 sin^{3} (x) - 5 sin^{2} (x) + 2 sin (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(x)+5sin(x)-4=0 ?
The general solution for 3cos^2(x)+5sin(x)-4=0 is x=0.23455…+2pin,x=pi-0.23455…+2pin