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

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

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

Solution

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

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

- 3

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Subtract the numbers:

3 - 1 = 2

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 3, c = 2

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

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

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

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 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 \cdot 1}

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

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

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

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = - 1, u = - 2

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = - 2

sin (x) = - 1, sin (x) = - 2

sin (x) = - 1 : x = \frac{3 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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Popular Examples

csc(4x)=0

csc (4 x) = 0

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2 sin (2 x) = - cos (2 x)

cos(2x)+cos(pi/2-2x)=-sqrt(2)

cos (2 x) + cos (\frac{π}{2} - 2 x) = - 2

cos^2(x)=-1.5

cos^{2} (x) = - 1.5

2arctan(1/2)-arctan(x)= pi/4

2 arctan (\frac{1}{2}) - arctan (x) = \frac{π}{4}

Frequently Asked Questions (FAQ)

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