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

The general solution for 3sin^2(x)=cos(x) is x=0.56024…+2pin,x=2pi-0.56024…+2pin

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Popular >

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

Solution

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

Solution

x = 0.56024 \dots + 2 πn, x = 2 π - 0.56024 \dots + 2 πn

Degrees

x = 32.09944 \dots^{\circ} + 36 0^{\circ} n, x = 327.90055 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

cos (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

- u + (1 - u^{2}) \cdot 3 = 0 : u = - \frac{1 + 37}{6}, u = \frac{37 - 1}{6}

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

Expand

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

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

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

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}

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 3) \cdot 3 = 37

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 3 \cdot 3 = 36

4 \cdot 3 \cdot 3

Multiply the numbers:

4 \cdot 3 \cdot 3 = 36

= 36

= 1 + 36

Add the numbers:

1 + 36 = 37

= 37

u_{1, 2} = \frac{- ( - 1 ) \pm 37}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 37}{2 ( - 3 )}, u_{2} = \frac{- ( - 1 ) - 37}{2 ( - 3 )}

u = \frac{- ( - 1 ) + 37}{2 ( - 3 )} : - \frac{1 + 37}{6}

\frac{- ( - 1 ) + 37}{2 ( - 3 )}

Remove parentheses:

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

= \frac{1 + 37}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{1 + 37}{- 6}

Apply the fraction rule:

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

= - \frac{1 + 37}{6}

u = \frac{- ( - 1 ) - 37}{2 ( - 3 )} : \frac{37 - 1}{6}

\frac{- ( - 1 ) - 37}{2 ( - 3 )}

Remove parentheses:

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

= \frac{1 - 37}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{1 - 37}{- 6}

Apply the fraction rule:

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

1 - 37 = - (37 - 1)

= \frac{37 - 1}{6}

The solutions to the quadratic equation are:

u = - \frac{1 + 37}{6}, u = \frac{37 - 1}{6}

Substitute back

u = cos (x)

cos (x) = - \frac{1 + 37}{6}, cos (x) = \frac{37 - 1}{6}

cos (x) = - \frac{1 + 37}{6}, cos (x) = \frac{37 - 1}{6}

cos (x) = - \frac{1 + 37}{6} :

No Solution

cos (x) = - \frac{1 + 37}{6}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{37 - 1}{6} : x = arccos (\frac{37 - 1}{6}) + 2 πn, x = 2 π - arccos (\frac{37 - 1}{6}) + 2 πn

cos (x) = \frac{37 - 1}{6}

Apply trig inverse properties

cos (x) = \frac{37 - 1}{6}

General solutions for

cos (x) = \frac{37 - 1}{6}

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

x = arccos (\frac{37 - 1}{6}) + 2 πn, x = 2 π - arccos (\frac{37 - 1}{6}) + 2 πn

x = arccos (\frac{37 - 1}{6}) + 2 πn, x = 2 π - arccos (\frac{37 - 1}{6}) + 2 πn

Combine all the solutions

x = arccos (\frac{37 - 1}{6}) + 2 πn, x = 2 π - arccos (\frac{37 - 1}{6}) + 2 πn

Show solutions in decimal form

x = 0.56024 \dots + 2 πn, x = 2 π - 0.56024 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(x)=cos(x) ?
The general solution for 3sin^2(x)=cos(x) is x=0.56024…+2pin,x=2pi-0.56024…+2pin