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

The general solution for 3cos(x)=2sin^2(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2 sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Expand

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

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

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

Expand

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

- 2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 3 + 5 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 3 - 5}{2 \cdot 2} : - 2

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

Subtract the numbers:

- 3 - 5 = - 8

= \frac{- 8}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - 2

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = - 2

cos (x) = \frac{1}{2}, cos (x) = - 2

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

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4 sin (θ) = 2

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6 cos^{2} (x) = 3

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3 sin (x) + 3 = sin (x)

4sec^2(x)-7tan^2(x)=3

4 sec^{2} (x) - 7 tan^{2} (x) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 3cos(x)=2sin^2(x) ?
The general solution for 3cos(x)=2sin^2(x) is x= pi/3+2pin,x=(5pi)/3+2pin