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Popular >

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

Solution

cos^{2} (x) + 3 ∣ cos (x) ∣ - 1 = 0

Solution

x = 1.87839 \dots + 2 πn, x = - 1.87839 \dots + 2 πn, x = 1.26319 \dots + 2 πn, x = 2 π - 1.26319 \dots + 2 πn

Degrees

x = 107.62439 \dots^{\circ} + 36 0^{\circ} n, x = - 107.62439 \dots^{\circ} + 36 0^{\circ} n, x = 72.37560 \dots^{\circ} + 36 0^{\circ} n, x = 287.62439 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) + 3 ∣ cos (x) ∣ - 1 = 0

Solve by substitution

cos^{2} (x) + 3 ∣ cos (x) ∣ - 1 = 0

Let:

cos (x) = u

u^{2} + 3 ∣ u ∣ - 1 = 0

u^{2} + 3 ∣ u ∣ - 1 = 0 : u = \frac{3 - 13}{2} or u = \frac{- 3 + 13}{2}

u^{2} + 3 ∣ u ∣ - 1 = 0

Find positive and negative intervals

Find intervals for

∣ u ∣

u \geq 0

u \geq 0, ∣ u ∣ = u

Rewrite

∣ u ∣

for

u \geq 0 : ∣ u ∣ = u

Apply absolute rule: If

u \geq 0

then

∣ u ∣ = u

∣ u ∣ = u

u < 0

u < 0, ∣ u ∣ = - u

Rewrite

∣ u ∣

for

u < 0 : ∣ u ∣ = - u

Apply absolute rule: If

u < 0

then

∣ u ∣ = - u

∣ u ∣ = - u

Identify the intervals:

u < 0, u \geq 0

∣ u ∣ u < 0 - u \geq 0 +

u < 0, u \geq 0

u < 0, u \geq 0

Solve the inequality for each interval

u < 0, u \geq 0

For

u < 0 : u = \frac{3 - 13}{2}

For

u < 0

rewrite

u^{2} + 3 ∣ u ∣ - 1 = 0

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

u^{2} + 3 (- u) - 1 = 0 : u = \frac{3 + 13}{2}, u = \frac{3 - 13}{2}

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

Refine

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}

Combine the intervals

(u = \frac{3 - 13}{2} or u = \frac{3 + 13}{2}) and (u < 0)

Merge Overlapping Intervals

u = \frac{3 - 13}{2} or u = \frac{3 + 13}{2} and u < 0

The intersection of two intervals is the set of numbers which are in both intervals

u = \frac{3 - 13}{2} or u = \frac{3 + 13}{2}

and

u < 0

u = \frac{3 - 13}{2}

u = \frac{3 - 13}{2}

For

u \geq 0 : u = \frac{- 3 + 13}{2}

For

u \geq 0

rewrite

u^{2} + 3 ∣ u ∣ - 1 = 0

u^{2} + 3 u - 1 = 0

u^{2} + 3 u - 1 = 0 : u = \frac{- 3 + 13}{2}, u = \frac{- 3 - 13}{2}

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

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}

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}

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}

Combine the intervals

(u = \frac{- 3 - 13}{2} or u = \frac{- 3 + 13}{2}) and (u \geq 0)

Merge Overlapping Intervals

u = \frac{- 3 - 13}{2} or u = \frac{- 3 + 13}{2} and u \geq 0

The intersection of two intervals is the set of numbers which are in both intervals

u = \frac{- 3 - 13}{2} or u = \frac{- 3 + 13}{2}

and

u \geq 0

u = \frac{- 3 + 13}{2}

u = \frac{- 3 + 13}{2}

Combine Solutions:

u = \frac{3 - 13}{2} or u = \frac{- 3 + 13}{2}

u = \frac{3 - 13}{2} or u = \frac{- 3 + 13}{2}

Substitute back

u = cos (x)

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

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

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

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

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

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

Show solutions in decimal form

x = 1.87839 \dots + 2 πn, x = - 1.87839 \dots + 2 πn, x = 1.26319 \dots + 2 πn, x = 2 π - 1.26319 \dots + 2 πn

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