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

The general solution for cos(2x)-3sin(-x)+2=0 is x=-0.75617…+2pin,x=pi+0.75617…+2pin

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

Solution

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

Solution

x = - 0.75617 \dots + 2 πn, x = π + 0.75617 \dots + 2 πn

Degrees

x = - 43.32537 \dots^{\circ} + 36 0^{\circ} n, x = 223.32537 \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))

Apply rule

- (- a) = a

= 2 + cos (2 x) + 3 sin (x)

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

Use the Double Angle identity:

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

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

Simplify

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

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

Solve by substitution

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

Let:

sin (x) = u

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

3 - 2 u^{2} + 3 u = 0 : u = - \frac{- 3 + 33}{4}, u = \frac{3 + 33}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

3^{2} - 4 (- 2) \cdot 3 = 33

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

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 3^{2} + 24

3^{2} = 9

= 9 + 24

Add the numbers:

9 + 24 = 33

= 33

u_{1, 2} = \frac{- 3 \pm 33}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 3 + 33}{2 ( - 2 )}, u_{2} = \frac{- 3 - 33}{2 ( - 2 )}

u = \frac{- 3 + 33}{2 ( - 2 )} : - \frac{- 3 + 33}{4}

\frac{- 3 + 33}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 33}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 3 + 33}{- 4}

Apply the fraction rule:

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

= - \frac{- 3 + 33}{4}

u = \frac{- 3 - 33}{2 ( - 2 )} : \frac{3 + 33}{4}

\frac{- 3 - 33}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 - 33}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 3 - 33}{- 4}

Apply the fraction rule:

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

- 3 - 33 = - (3 + 33)

= \frac{3 + 33}{4}

The solutions to the quadratic equation are:

u = - \frac{- 3 + 33}{4}, u = \frac{3 + 33}{4}

Substitute back

u = sin (x)

sin (x) = - \frac{- 3 + 33}{4}, sin (x) = \frac{3 + 33}{4}

sin (x) = - \frac{- 3 + 33}{4}, sin (x) = \frac{3 + 33}{4}

sin (x) = - \frac{- 3 + 33}{4} : x = arcsin (- \frac{- 3 + 33}{4}) + 2 πn, x = π + arcsin (\frac{- 3 + 33}{4}) + 2 πn

sin (x) = - \frac{- 3 + 33}{4}

Apply trig inverse properties

sin (x) = - \frac{- 3 + 33}{4}

General solutions for

sin (x) = - \frac{- 3 + 33}{4}

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

x = arcsin (- \frac{- 3 + 33}{4}) + 2 πn, x = π + arcsin (\frac{- 3 + 33}{4}) + 2 πn

x = arcsin (- \frac{- 3 + 33}{4}) + 2 πn, x = π + arcsin (\frac{- 3 + 33}{4}) + 2 πn

sin (x) = \frac{3 + 33}{4} :

No Solution

sin (x) = \frac{3 + 33}{4}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- \frac{- 3 + 33}{4}) + 2 πn, x = π + arcsin (\frac{- 3 + 33}{4}) + 2 πn

Show solutions in decimal form

x = - 0.75617 \dots + 2 πn, x = π + 0.75617 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)-3sin(-x)+2=0 ?
The general solution for cos(2x)-3sin(-x)+2=0 is x=-0.75617…+2pin,x=pi+0.75617…+2pin