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

The general solution for 2cos^2(x)-3sin(-x)-3=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x= pi/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

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

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

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

Simplify

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

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

Apply rule

- (- a) = a

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

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Add/Subtract the numbers:

- 3 + 2 = - 1

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 3 - 1 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

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

Popular Examples

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

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

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

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

-2sin(x)=sqrt(3)

- 2 sin (x) = 3

((sqrt(2))/2)(cos(θ)+sin(θ))=1

(\frac{2}{2}) (cos (θ) + sin (θ)) = 1

-6cos^2(θ)+14cos(θ)+15=3cos(θ)+8

- 6 cos^{2} (θ) + 14 cos (θ) + 15 = 3 cos (θ) + 8

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)-3sin(-x)-3=0 ?
The general solution for 2cos^2(x)-3sin(-x)-3=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x= pi/2+2pin