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

The general solution for cos^2(x)-sin^2(x)=cos^2(x)+3cos(x) is x=1.87839…+2pin,x=-1.87839…+2pin

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

Solution

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

Solution

x = 1.87839 \dots + 2 πn, x = - 1.87839 \dots + 2 πn

Degrees

x = 107.62439 \dots^{\circ} + 36 0^{\circ} n, x = - 107.62439 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

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

Solve by substitution

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

Let:

cos (x) = 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

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 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}

Apply rule

- (- a) = a

= \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}

Apply rule

- (- a) = a

= \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 (x)

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

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

cos (x) = \frac{3 + 13}{2} :

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 1.87839 \dots + 2 πn, x = - 1.87839 \dots + 2 πn

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cos (x) = - 0, 4

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tan (a) = 0.5

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sin(x)-cos(x)=-2

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-sin^2(x)=cos^2(x)+3cos(x) ?
The general solution for cos^2(x)-sin^2(x)=cos^2(x)+3cos(x) is x=1.87839…+2pin,x=-1.87839…+2pin