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Popular >

4cos^2(x)+sqrt(3)=2(sqrt(3)+1)

Solution

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

Solution

x = 0.26179 \dots + 2 πn, x = 2 π - 0.26179 \dots + 2 πn, x = 2.87979 \dots + 2 πn, x = - 2.87979 \dots + 2 πn

Degrees

x = 1 5^{\circ} + 36 0^{\circ} n, x = 34 5^{\circ} + 36 0^{\circ} n, x = 16 5^{\circ} + 36 0^{\circ} n, x = - 16 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

4 u^{2} + 3 = 2 (3 + 1) : u = \frac{3 + 2}{2}, u = - \frac{3 + 2}{2}

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

Move

3

to the right side

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

Subtract

3

from both sides

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

Simplify

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

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

Divide both sides by

4

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

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 u ^{2}}{4} : u^{2}

\frac{4 u ^{2}}{4}

Divide the numbers:

\frac{4}{4} = 1

= u^{2}

Simplify

\frac{2 ( 3 + 1 )}{4} - \frac{3}{4} : \frac{3 + 2}{4}

\frac{2 ( 3 + 1 )}{4} - \frac{3}{4}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Expand

2 (3 + 1) - 3 : 3 + 2

2 (3 + 1) - 3

Expand

2 (3 + 1) : 23 + 2

2 (3 + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = 3, c = 1

= 23 + 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 23 + 2

= 23 + 2 - 3

Add similar elements:

23 - 3 = 3

= 3 + 2

= \frac{3 + 2}{4}

u^{2} = \frac{3 + 2}{4}

u^{2} = \frac{3 + 2}{4}

u^{2} = \frac{3 + 2}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{3 + 2}{4}, u = - \frac{3 + 2}{4}

\frac{3 + 2}{4} = \frac{3 + 2}{2}

\frac{3 + 2}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3 + 2}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3 + 2}{2}

- \frac{3 + 2}{4} = - \frac{3 + 2}{2}

- \frac{3 + 2}{4}

Simplify

\frac{3 + 2}{4} : \frac{3 + 2}{2}

\frac{3 + 2}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3 + 2}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3 + 2}{2}

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

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

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

Substitute back

u = cos (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 0.26179 \dots + 2 πn, x = 2 π - 0.26179 \dots + 2 πn, x = 2.87979 \dots + 2 πn, x = - 2.87979 \dots + 2 πn

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