What is the general solution for 7cos(x)+3=3cos(x)+2 ?

The general solution for 7cos(x)+3=3cos(x)+2 is x=1.82347…+2pin,x=-1.82347…+2pin

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7cos(x)+3=3cos(x)+2

7 cos (x) + 3 = 3 cos (x) + 2

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

Degrees

x = 104.47751 \dots^{\circ} + 36 0^{\circ} n, x = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

7 cos (x) + 3 = 3 cos (x) + 2

Solve by substitution

7 cos (x) + 3 = 3 cos (x) + 2

Let:

cos (x) = u

7 u + 3 = 3 u + 2

7 u + 3 = 3 u + 2 : u = - \frac{1}{4}

7 u + 3 = 3 u + 2

Move

3

to the right side

7 u + 3 = 3 u + 2

Subtract

3

from both sides

7 u + 3 - 3 = 3 u + 2 - 3

Simplify

7 u = 3 u - 1

7 u = 3 u - 1

Move

3 u

to the left side

7 u = 3 u - 1

Subtract

3 u

from both sides

7 u - 3 u = 3 u - 1 - 3 u

Simplify

4 u = - 1

4 u = - 1

Divide both sides by

4

4 u = - 1

Divide both sides by

4

\frac{4 u}{4} = \frac{- 1}{4}

Simplify

u = - \frac{1}{4}

u = - \frac{1}{4}

Substitute back

u = cos (x)

cos (x) = - \frac{1}{4}

cos (x) = - \frac{1}{4}

cos (x) = - \frac{1}{4} : x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

cos (x) = - \frac{1}{4}

Apply trig inverse properties

cos (x) = - \frac{1}{4}

General solutions for

cos (x) = - \frac{1}{4}

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

x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

Combine all the solutions

x = arccos (- \frac{1}{4}) + 2 πn, x = - arccos (- \frac{1}{4}) + 2 πn

Show solutions in decimal form

x = 1.82347 \dots + 2 πn, x = - 1.82347 \dots + 2 πn

2 sin (2 θ) + 1 = 0

sin (2 x) - 2 sin (x) = 0

7 cos^{2} (x) - 1 = 5 cos (x)

x, sin (2 x) = 0

cot (θ) = - 4

What is the general solution for 7cos(x)+3=3cos(x)+2 ?
The general solution for 7cos(x)+3=3cos(x)+2 is x=1.82347…+2pin,x=-1.82347…+2pin