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

The general solution for 2sin^2(x)+5cos(x)-3=0 is x=1.34977…+2pin,x=2pi-1.34977…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = 1.34977 \dots + 2 πn, x = 2 π - 1.34977 \dots + 2 πn

Degrees

x = 77.33656 \dots^{\circ} + 36 0^{\circ} n, x = 282.66343 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

2 (1 - cos^{2} (x)) : 2 - 2 cos^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Add/Subtract the numbers:

- 3 + 2 = - 1

= 5 cos (x) - 2 cos^{2} (x) - 1

= 5 cos (x) - 2 cos^{2} (x) - 1

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

Solve by substitution

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

Let:

cos (x) = u

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

- 1 - 2 u^{2} + 5 u = 0 : u = \frac{5 - 17}{4}, u = \frac{5 + 17}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = 5, c = - 1

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

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

5^{2} - 4 (- 2) (- 1) = 17

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 5^{2} - 8

5^{2} = 25

= 25 - 8

Subtract the numbers:

25 - 8 = 17

= 17

u_{1, 2} = \frac{- 5 \pm 17}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 5 + 17}{2 ( - 2 )}, u_{2} = \frac{- 5 - 17}{2 ( - 2 )}

u = \frac{- 5 + 17}{2 ( - 2 )} : \frac{5 - 17}{4}

\frac{- 5 + 17}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 5 + 17}{- 4}

Apply the fraction rule:

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

- 5 + 17 = - (5 - 17)

= \frac{5 - 17}{4}

u = \frac{- 5 - 17}{2 ( - 2 )} : \frac{5 + 17}{4}

\frac{- 5 - 17}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 5 - 17}{- 4}

Apply the fraction rule:

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

- 5 - 17 = - (5 + 17)

= \frac{5 + 17}{4}

The solutions to the quadratic equation are:

u = \frac{5 - 17}{4}, u = \frac{5 + 17}{4}

Substitute back

u = cos (x)

cos (x) = \frac{5 - 17}{4}, cos (x) = \frac{5 + 17}{4}

cos (x) = \frac{5 - 17}{4}, cos (x) = \frac{5 + 17}{4}

cos (x) = \frac{5 - 17}{4} : x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

cos (x) = \frac{5 - 17}{4}

Apply trig inverse properties

cos (x) = \frac{5 - 17}{4}

General solutions for

cos (x) = \frac{5 - 17}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

cos (x) = \frac{5 + 17}{4} :

No Solution

cos (x) = \frac{5 + 17}{4}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

Show solutions in decimal form

x = 1.34977 \dots + 2 πn, x = 2 π - 1.34977 \dots + 2 πn

Popular Examples

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

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

sin(θ)=-27/190

sin (θ) = - \frac{27}{190}

4sin(θ)=1+sin(θ)

4 sin (θ) = 1 + sin (θ)

tan(θ)-sqrt(3)=0,0<= θ<= 2pi

tan (θ) - 3 = 0, 0 \leq θ \leq 2 π

1/32*1200^2*sin(2θ)=3000

\frac{1}{32} \cdot 120 0^{2} \cdot sin (2 θ) = 3000

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)+5cos(x)-3=0 ?
The general solution for 2sin^2(x)+5cos(x)-3=0 is x=1.34977…+2pin,x=2pi-1.34977…+2pin