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

The general solution for cos^2(x)+3sin(-x)-2=0 is x=-0.39192…+2pin,x=pi+0.39192…+2pin

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Popular >

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

Solution

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

Solution

x = - 0.39192 \dots + 2 πn, x = π + 0.39192 \dots + 2 πn

Degrees

x = - 22.45551 \dots^{\circ} + 36 0^{\circ} n, x = 202.45551 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

- 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)

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

Simplify

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 3^{2} - 4

3^{2} = 9

= 9 - 4

Subtract the numbers:

9 - 4 = 5

= 5

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

Separate the solutions

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

u = \frac{- ( - 3 ) + 5}{2 ( - 1 )} : - \frac{3 + 5}{2}

\frac{- ( - 3 ) + 5}{2 ( - 1 )}

Remove parentheses:

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

= \frac{3 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

u = \frac{- ( - 3 ) - 5}{2 ( - 1 )} : - \frac{3 - 5}{2}

\frac{- ( - 3 ) - 5}{2 ( - 1 )}

Remove parentheses:

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

= \frac{3 - 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

The solutions to the quadratic equation are:

u = - \frac{3 + 5}{2}, u = - \frac{3 - 5}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{3 + 5}{2}, sin (x) = - \frac{3 - 5}{2}

sin (x) = - \frac{3 + 5}{2}, sin (x) = - \frac{3 - 5}{2}

sin (x) = - \frac{3 + 5}{2} :

No Solution

sin (x) = - \frac{3 + 5}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - \frac{3 - 5}{2} : x = arcsin (- \frac{3 - 5}{2}) + 2 πn, x = π + arcsin (\frac{3 - 5}{2}) + 2 πn

sin (x) = - \frac{3 - 5}{2}

Apply trig inverse properties

sin (x) = - \frac{3 - 5}{2}

General solutions for

sin (x) = - \frac{3 - 5}{2}

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

x = arcsin (- \frac{3 - 5}{2}) + 2 πn, x = π + arcsin (\frac{3 - 5}{2}) + 2 πn

x = arcsin (- \frac{3 - 5}{2}) + 2 πn, x = π + arcsin (\frac{3 - 5}{2}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{3 - 5}{2}) + 2 πn, x = π + arcsin (\frac{3 - 5}{2}) + 2 πn

Show solutions in decimal form

x = - 0.39192 \dots + 2 πn, x = π + 0.39192 \dots + 2 πn

Popular Examples

cos(x)=-1,0<x<2pi

cos (x) = - 1, 0 < x < 2 π

1/(1+cos(x))+1/(1-cos(x))=32

\frac{1}{1 + cos ( x )} + \frac{1}{1 - cos ( x )} = 32

sin^2(x)=14

sin^{2} (x) = 14

sin(2x)*cos(3x)=0

sin (2 x) \cdot cos (3 x) = 0

sin^2(x)=-3

sin^{2} (x) = - 3

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)+3sin(-x)-2=0 ?
The general solution for cos^2(x)+3sin(-x)-2=0 is x=-0.39192…+2pin,x=pi+0.39192…+2pin