What is the general solution for 2+3cos^2(x)=sin^2(x)+4cos(x) ?

The general solution for 2+3cos^2(x)=sin^2(x)+4cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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2+3cos^2(x)=sin^2(x)+4cos(x)

Solution

2 + 3 cos^{2} (x) = sin^{2} (x) + 4 cos (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

2 + 3 cos^{2} (x) = sin^{2} (x) + 4 cos (x)

Subtract

sin^{2} (x) + 4 cos (x)

from both sides

2 + 3 cos^{2} (x) - sin^{2} (x) - 4 cos (x) = 0

Rewrite using trig identities

2 - sin^{2} (x) + 3 cos^{2} (x) - 4 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^{2} (x) - 4 cos (x)

Simplify

2 - (1 - cos^{2} (x)) + 3 cos^{2} (x) - 4 cos (x) : 4 cos^{2} (x) - 4 cos (x) + 1

2 - (1 - cos^{2} (x)) + 3 cos^{2} (x) - 4 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)

= 2 - 1 + cos^{2} (x) + 3 cos^{2} (x) - 4 cos (x)

Simplify

2 - 1 + cos^{2} (x) + 3 cos^{2} (x) - 4 cos (x) : 4 cos^{2} (x) - 4 cos (x) + 1

2 - 1 + cos^{2} (x) + 3 cos^{2} (x) - 4 cos (x)

Add similar elements:

cos^{2} (x) + 3 cos^{2} (x) = 4 cos^{2} (x)

= 2 - 1 + 4 cos^{2} (x) - 4 cos (x)

Subtract the numbers:

2 - 1 = 1

= 4 cos^{2} (x) - 4 cos (x) + 1

= 4 cos^{2} (x) - 4 cos (x) + 1

= 4 cos^{2} (x) - 4 cos (x) + 1

1 - 4 cos (x) + 4 cos^{2} (x) = 0

Solve by substitution

1 - 4 cos (x) + 4 cos^{2} (x) = 0

Let:

cos (x) = u

1 - 4 u + 4 u^{2} = 0

1 - 4 u + 4 u^{2} = 0 : u = \frac{1}{2}

1 - 4 u + 4 u^{2} = 0

Write in the standard form

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

4 u^{2} - 4 u + 1 = 0

Solve with the quadratic formula

4 u^{2} - 4 u + 1 = 0

Quadratic Equation Formula:

For

a = 4, b = - 4, c = 1

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

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

(- 4)^{2} - 4 \cdot 4 \cdot 1 = 0

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 4}

u = \frac{- ( - 4 )}{2 \cdot 4}

\frac{- ( - 4 )}{2 \cdot 4} = \frac{1}{2}

\frac{- ( - 4 )}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{4}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{1}{2}

The solution to the quadratic equation is:

u = \frac{1}{2}

Substitute back

u = cos (x)

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

cos (x) = \frac{1}{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

Combine all the solutions

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

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tan(4x)-tan(2x)=0

tan (4 x) - tan (2 x) = 0

solvefor t,sin(t)+2sin(2t)=0 solve for

t, sin (t) + 2 sin (2 t) = 0

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-1-tan(x)=sec(x),0<= x<2pi

- 1 - tan (x) = sec (x), 0 \leq x < 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 2+3cos^2(x)=sin^2(x)+4cos(x) ?
The general solution for 2+3cos^2(x)=sin^2(x)+4cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin