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

The general solution for 3cos(A)=sin^2(A) is A=1.26319…+2pin,A=2pi-1.26319…+2pin

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

Solution

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

Solution

A = 1.26319 \dots + 2 πn, A = 2 π - 1.26319 \dots + 2 πn

Degrees

A = 72.37560 \dots^{\circ} + 36 0^{\circ} n, A = 287.62439 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

sin^{2} (A)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

- (1 - cos^{2} (A)) : - 1 + cos^{2} (A)

- (1 - cos^{2} (A))

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (A)

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

- 1 + cos^{2} (A) + 3 cos (A) = 0

Solve by substitution

- 1 + cos^{2} (A) + 3 cos (A) = 0

Let:

cos (A) = u

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

- 1 + u^{2} + 3 u = 0 : u = \frac{- 3 + 13}{2}, u = \frac{- 3 - 13}{2}

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

Write in the standard form

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

u^{2} + 3 u - 1 = 0

Solve with the quadratic formula

u^{2} + 3 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 3, c = - 1

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

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

3^{2} - 4 \cdot 1 \cdot (- 1) = 13

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

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 3^{2} + 4

3^{2} = 9

= 9 + 4

Add the numbers:

9 + 4 = 13

= 13

u_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 3 + 13}{2 \cdot 1}, u_{2} = \frac{- 3 - 13}{2 \cdot 1}

u = \frac{- 3 + 13}{2 \cdot 1} : \frac{- 3 + 13}{2}

\frac{- 3 + 13}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 3 + 13}{2}

u = \frac{- 3 - 13}{2 \cdot 1} : \frac{- 3 - 13}{2}

\frac{- 3 - 13}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 3 - 13}{2}

The solutions to the quadratic equation are:

u = \frac{- 3 + 13}{2}, u = \frac{- 3 - 13}{2}

Substitute back

u = cos (A)

cos (A) = \frac{- 3 + 13}{2}, cos (A) = \frac{- 3 - 13}{2}

cos (A) = \frac{- 3 + 13}{2}, cos (A) = \frac{- 3 - 13}{2}

cos (A) = \frac{- 3 + 13}{2} : A = arccos (\frac{- 3 + 13}{2}) + 2 πn, A = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

cos (A) = \frac{- 3 + 13}{2}

Apply trig inverse properties

cos (A) = \frac{- 3 + 13}{2}

General solutions for

cos (A) = \frac{- 3 + 13}{2}

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

A = arccos (\frac{- 3 + 13}{2}) + 2 πn, A = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

A = arccos (\frac{- 3 + 13}{2}) + 2 πn, A = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

cos (A) = \frac{- 3 - 13}{2} :

No Solution

cos (A) = \frac{- 3 - 13}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

A = arccos (\frac{- 3 + 13}{2}) + 2 πn, A = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

Show solutions in decimal form

A = 1.26319 \dots + 2 πn, A = 2 π - 1.26319 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos(A)=sin^2(A) ?
The general solution for 3cos(A)=sin^2(A) is A=1.26319…+2pin,A=2pi-1.26319…+2pin