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Popular >

sqrt(4cos^2(x)-1)=1

Solution

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

Solution

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

4 u^{2} - 1 = 1

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

4 u^{2} - 1 = 1

Square both sides:

4 u^{2} - 1 = 1

4 u^{2} - 1 = 1

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

Expand

(4 u^{2} - 1)^{2} : 4 u^{2} - 1

(4 u^{2} - 1)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 4 u^{2} - 1

Expand

1^{2} : 1

1^{2}

Apply rule

1^{a} = 1

= 1

4 u^{2} - 1 = 1

4 u^{2} - 1 = 1

Solve

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

4 u^{2} - 1 = 1

Move

1

to the right side

4 u^{2} - 1 = 1

Add

1

to both sides

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

Simplify

4 u^{2} = 2

4 u^{2} = 2

Divide both sides by

4

4 u^{2} = 2

Divide both sides by

4

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

Simplify

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{2}, u = - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

Verify Solutions:

u = \frac{1}{2}

True

, u = - \frac{1}{2}

True

Check the solutions by plugging them into

4 u^{2} - 1 = 1

Remove the ones that don't agree with the equation.

Plug in

u = \frac{1}{2} :

True

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

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

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

4 (\frac{1}{2})^{2} = 2

4 (\frac{1}{2})^{2}

(\frac{1}{2})^{2} = \frac{1}{2}

(\frac{1}{2})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((\frac{1}{2})^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (\frac{1}{2})^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= \frac{1}{2}

= 4 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 4}{2}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

Apply rule

1 = 1

= 1

1 = 1

True

Plug in

u = - \frac{1}{2} :

True

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

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

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

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

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

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

(- \frac{1}{2})^{2}

Apply exponent rule:

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

n

is even

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

= (\frac{1}{2})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((\frac{1}{2})^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (\frac{1}{2})^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= \frac{1}{2}

= 4 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 4}{2}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

= 2 - 1

Subtract the numbers:

2 - 1 = 1

= 1

Apply rule

1 = 1

= 1

1 = 1

True

The solutions are

u = \frac{1}{2}, u = - \frac{1}{2}

Substitute back

u = cos (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

Graph

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