What is the general solution for 3cos(2x)-5cos(x)=0 ?

The general solution for 3cos(2x)-5cos(x)=0 is x=1.98675…+2pin,x=-1.98675…+2pin

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3cos(2x)-5cos(x)=0

Solution

3 cos (2 x) - 5 cos (x) = 0

Solution

x = 1.98675 \dots + 2 πn, x = - 1.98675 \dots + 2 πn

Degrees

x = 113.83295 \dots^{\circ} + 36 0^{\circ} n, x = - 113.83295 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (2 x) - 5 cos (x) = 0

Rewrite using trig identities

3 cos (2 x) - 5 cos (x)

Use the Double Angle identity:

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

= 3 (2 cos^{2} (x) - 1) - 5 cos (x)

(- 1 + 2 cos^{2} (x)) \cdot 3 - 5 cos (x) = 0

Solve by substitution

(- 1 + 2 cos^{2} (x)) \cdot 3 - 5 cos (x) = 0

Let:

cos (x) = u

(- 1 + 2 u^{2}) \cdot 3 - 5 u = 0

(- 1 + 2 u^{2}) \cdot 3 - 5 u = 0 : u = \frac{5 + 97}{12}, u = \frac{5 - 97}{12}

(- 1 + 2 u^{2}) \cdot 3 - 5 u = 0

Expand

(- 1 + 2 u^{2}) \cdot 3 - 5 u : - 3 + 6 u^{2} - 5 u

(- 1 + 2 u^{2}) \cdot 3 - 5 u

= 3 (- 1 + 2 u^{2}) - 5 u

Expand

3 (- 1 + 2 u^{2}) : - 3 + 6 u^{2}

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

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = - 1, c = 2 u^{2}

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

Apply minus-plus rules

+ (- a) = - a

= - 3 \cdot 1 + 3 \cdot 2 u^{2}

Simplify

- 3 \cdot 1 + 3 \cdot 2 u^{2} : - 3 + 6 u^{2}

- 3 \cdot 1 + 3 \cdot 2 u^{2}

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 \cdot 2 u^{2}

Multiply the numbers:

3 \cdot 2 = 6

= - 3 + 6 u^{2}

= - 3 + 6 u^{2}

= - 3 + 6 u^{2} - 5 u

- 3 + 6 u^{2} - 5 u = 0

Write in the standard form

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

6 u^{2} - 5 u - 3 = 0

Solve with the quadratic formula

6 u^{2} - 5 u - 3 = 0

Quadratic Equation Formula:

For

a = 6, b = - 5, c = - 3

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 (- 3) = 97

(- 5)^{2} - 4 \cdot 6 (- 3)

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 6 \cdot 3

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 6 \cdot 3

Multiply the numbers:

4 \cdot 6 \cdot 3 = 72

= 5^{2} + 72

5^{2} = 25

= 25 + 72

Add the numbers:

25 + 72 = 97

= 97

u_{1, 2} = \frac{- ( - 5 ) \pm 97}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 97}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 97}{2 \cdot 6}

u = \frac{- ( - 5 ) + 97}{2 \cdot 6} : \frac{5 + 97}{12}

\frac{- ( - 5 ) + 97}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 + 97}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{5 + 97}{12}

u = \frac{- ( - 5 ) - 97}{2 \cdot 6} : \frac{5 - 97}{12}

\frac{- ( - 5 ) - 97}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 - 97}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{5 - 97}{12}

The solutions to the quadratic equation are:

u = \frac{5 + 97}{12}, u = \frac{5 - 97}{12}

Substitute back

u = cos (x)

cos (x) = \frac{5 + 97}{12}, cos (x) = \frac{5 - 97}{12}

cos (x) = \frac{5 + 97}{12}, cos (x) = \frac{5 - 97}{12}

cos (x) = \frac{5 + 97}{12} :

No Solution

cos (x) = \frac{5 + 97}{12}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{5 - 97}{12} : x = arccos (\frac{5 - 97}{12}) + 2 πn, x = - arccos (\frac{5 - 97}{12}) + 2 πn

cos (x) = \frac{5 - 97}{12}

Apply trig inverse properties

cos (x) = \frac{5 - 97}{12}

General solutions for

cos (x) = \frac{5 - 97}{12}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (\frac{5 - 97}{12}) + 2 πn, x = - arccos (\frac{5 - 97}{12}) + 2 πn

x = arccos (\frac{5 - 97}{12}) + 2 πn, x = - arccos (\frac{5 - 97}{12}) + 2 πn

Combine all the solutions

x = arccos (\frac{5 - 97}{12}) + 2 πn, x = - arccos (\frac{5 - 97}{12}) + 2 πn

Show solutions in decimal form

x = 1.98675 \dots + 2 πn, x = - 1.98675 \dots + 2 πn

Popular Examples

sec(3θ)-2=0

sec (3 θ) - 2 = 0

sin(5x)=-1

sin (5 x) = - 1

tan^2(x)+3sec(x)=-3

tan^{2} (x) + 3 sec (x) = - 3

csc(x)=-(2sqrt(3))/3

csc (x) = - \frac{2 3}{3}

sin(2x)=-5cos(2x),0<= x<= 2pi

sin (2 x) = - 5 cos (2 x), 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 3cos(2x)-5cos(x)=0 ?
The general solution for 3cos(2x)-5cos(x)=0 is x=1.98675…+2pin,x=-1.98675…+2pin