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

The general solution for 2cos(a)=3sin^2(a) is a=0.76589…+2pin,a=2pi-0.76589…+2pin

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

Solution

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

Solution

a = 0.76589 \dots + 2 πn, a = 2 π - 0.76589 \dots + 2 πn

Degrees

a = 43.88280 \dots^{\circ} + 36 0^{\circ} n, a = 316.11719 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

3 sin^{2} (a)

from both sides

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

Rewrite using trig identities

2 cos (a) - 3 sin^{2} (a)

Use the Pythagorean identity:

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

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

= 2 cos (a) - 3 (1 - cos^{2} (a))

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

Solve by substitution

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

Let:

cos (a) = u

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

- (1 - u^{2}) \cdot 3 + 2 u = 0 : u = \frac{- 1 + 10}{3}, u = - \frac{1 + 10}{3}

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

Expand

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

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

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

Expand

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

- 3 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

2^{2} - 4 \cdot 3 (- 3) = 210

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 2^{2} + 36

2^{2} = 4

= 4 + 36

Add the numbers:

4 + 36 = 40

= 40

Prime factorization of

40 : 2^{3} \cdot 5

40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 5

= 2^{3} \cdot 5

= 2^{3} \cdot 5

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2 \cdot 5

Apply radical rule:

= 2^{2} 2 \cdot 5

Apply radical rule:

2^{2} = 2

= 2 2 \cdot 5

Refine

= 210

u_{1, 2} = \frac{- 2 \pm 2 10}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 2 + 2 10}{2 \cdot 3}, u_{2} = \frac{- 2 - 2 10}{2 \cdot 3}

u = \frac{- 2 + 2 10}{2 \cdot 3} : \frac{- 1 + 10}{3}

\frac{- 2 + 2 10}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2 + 2 10}{6}

Factor

- 2 + 210 : 2 (- 1 + 10)

- 2 + 210

Rewrite as

= - 2 \cdot 1 + 210

Factor out common term

2

= 2 (- 1 + 10)

= \frac{2 ( - 1 + 10 )}{6}

Cancel the common factor:

2

= \frac{- 1 + 10}{3}

u = \frac{- 2 - 2 10}{2 \cdot 3} : - \frac{1 + 10}{3}

\frac{- 2 - 2 10}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2 - 2 10}{6}

Factor

- 2 - 210 : - 2 (1 + 10)

- 2 - 210

Rewrite as

= - 2 \cdot 1 - 210

Factor out common term

2

= - 2 (1 + 10)

= - \frac{2 ( 1 + 10 )}{6}

Cancel the common factor:

2

= - \frac{1 + 10}{3}

The solutions to the quadratic equation are:

u = \frac{- 1 + 10}{3}, u = - \frac{1 + 10}{3}

Substitute back

u = cos (a)

cos (a) = \frac{- 1 + 10}{3}, cos (a) = - \frac{1 + 10}{3}

cos (a) = \frac{- 1 + 10}{3}, cos (a) = - \frac{1 + 10}{3}

cos (a) = \frac{- 1 + 10}{3} : a = arccos (\frac{- 1 + 10}{3}) + 2 πn, a = 2 π - arccos (\frac{- 1 + 10}{3}) + 2 πn

cos (a) = \frac{- 1 + 10}{3}

Apply trig inverse properties

cos (a) = \frac{- 1 + 10}{3}

General solutions for

cos (a) = \frac{- 1 + 10}{3}

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

a = arccos (\frac{- 1 + 10}{3}) + 2 πn, a = 2 π - arccos (\frac{- 1 + 10}{3}) + 2 πn

a = arccos (\frac{- 1 + 10}{3}) + 2 πn, a = 2 π - arccos (\frac{- 1 + 10}{3}) + 2 πn

cos (a) = - \frac{1 + 10}{3} :

No Solution

cos (a) = - \frac{1 + 10}{3}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

a = arccos (\frac{- 1 + 10}{3}) + 2 πn, a = 2 π - arccos (\frac{- 1 + 10}{3}) + 2 πn

Show solutions in decimal form

a = 0.76589 \dots + 2 πn, a = 2 π - 0.76589 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(a)=3sin^2(a) ?
The general solution for 2cos(a)=3sin^2(a) is a=0.76589…+2pin,a=2pi-0.76589…+2pin