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

The general solution for -3sin(x)-2cos(2x)=0 is x=-0.43939…+2pin,x=pi+0.43939…+2pin

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Popular >

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

Solution

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

Solution

x = - 0.43939 \dots + 2 πn, x = π + 0.43939 \dots + 2 πn

Degrees

x = - 25.17538 \dots^{\circ} + 36 0^{\circ} n, x = 205.17538 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Solve by substitution

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

Let:

sin (x) = u

- (1 - 2 u^{2}) \cdot 2 - 3 u = 0

- (1 - 2 u^{2}) \cdot 2 - 3 u = 0 : u = \frac{3 + 41}{8}, u = \frac{3 - 41}{8}

- (1 - 2 u^{2}) \cdot 2 - 3 u = 0

Expand

- (1 - 2 u^{2}) \cdot 2 - 3 u : - 2 + 4 u^{2} - 3 u

- (1 - 2 u^{2}) \cdot 2 - 3 u

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

Expand

- 2 (1 - 2 u^{2}) : - 2 + 4 u^{2}

- 2 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = 2 u^{2}

= - 2 \cdot 1 - (- 2) \cdot 2 u^{2}

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 \cdot 2 u^{2}

Simplify

- 2 \cdot 1 + 2 \cdot 2 u^{2} : - 2 + 4 u^{2}

- 2 \cdot 1 + 2 \cdot 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 \cdot 2 u^{2}

Multiply the numbers:

2 \cdot 2 = 4

= - 2 + 4 u^{2}

= - 2 + 4 u^{2}

= - 2 + 4 u^{2} - 3 u

- 2 + 4 u^{2} - 3 u = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 3, c = - 2

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

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

(- 3)^{2} - 4 \cdot 4 (- 2) = 41

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 3^{2} + 32

3^{2} = 9

= 9 + 32

Add the numbers:

9 + 32 = 41

= 41

u_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 4}, u_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 4}

u = \frac{- ( - 3 ) + 41}{2 \cdot 4} : \frac{3 + 41}{8}

\frac{- ( - 3 ) + 41}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{3 + 41}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3 + 41}{8}

u = \frac{- ( - 3 ) - 41}{2 \cdot 4} : \frac{3 - 41}{8}

\frac{- ( - 3 ) - 41}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{3 - 41}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3 - 41}{8}

The solutions to the quadratic equation are:

u = \frac{3 + 41}{8}, u = \frac{3 - 41}{8}

Substitute back

u = sin (x)

sin (x) = \frac{3 + 41}{8}, sin (x) = \frac{3 - 41}{8}

sin (x) = \frac{3 + 41}{8}, sin (x) = \frac{3 - 41}{8}

sin (x) = \frac{3 + 41}{8} :

No Solution

sin (x) = \frac{3 + 41}{8}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{3 - 41}{8} : x = arcsin (\frac{3 - 41}{8}) + 2 πn, x = π + arcsin (- \frac{3 - 41}{8}) + 2 πn

sin (x) = \frac{3 - 41}{8}

Apply trig inverse properties

sin (x) = \frac{3 - 41}{8}

General solutions for

sin (x) = \frac{3 - 41}{8}

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

x = arcsin (\frac{3 - 41}{8}) + 2 πn, x = π + arcsin (- \frac{3 - 41}{8}) + 2 πn

x = arcsin (\frac{3 - 41}{8}) + 2 πn, x = π + arcsin (- \frac{3 - 41}{8}) + 2 πn

Combine all the solutions

x = arcsin (\frac{3 - 41}{8}) + 2 πn, x = π + arcsin (- \frac{3 - 41}{8}) + 2 πn

Show solutions in decimal form

x = - 0.43939 \dots + 2 πn, x = π + 0.43939 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -3sin(x)-2cos(2x)=0 ?
The general solution for -3sin(x)-2cos(2x)=0 is x=-0.43939…+2pin,x=pi+0.43939…+2pin