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Popular >

12cos^2(x)-7cos(x)+1=0

Solution

12 cos^{2} (x) - 7 cos (x) + 1 = 0

Solution

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

Degrees

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n, x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

12 cos^{2} (x) - 7 cos (x) + 1 = 0

Solve by substitution

12 cos^{2} (x) - 7 cos (x) + 1 = 0

Let:

cos (x) = u

12 u^{2} - 7 u + 1 = 0

12 u^{2} - 7 u + 1 = 0 : u = \frac{1}{3}, u = \frac{1}{4}

12 u^{2} - 7 u + 1 = 0

Solve with the quadratic formula

12 u^{2} - 7 u + 1 = 0

Quadratic Equation Formula:

For

a = 12, b = - 7, c = 1

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

(- 7)^{2} - 4 \cdot 12 \cdot 1 = 1

(- 7)^{2} - 4 \cdot 12 \cdot 1

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 12 \cdot 1

Multiply the numbers:

4 \cdot 12 \cdot 1 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

Subtract the numbers:

49 - 48 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 12}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 12}, u_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 12}

u = \frac{- ( - 7 ) + 1}{2 \cdot 12} : \frac{1}{3}

\frac{- ( - 7 ) + 1}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{7 + 1}{2 \cdot 12}

Add the numbers:

7 + 1 = 8

= \frac{8}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{8}{24}

Cancel the common factor:

8

= \frac{1}{3}

u = \frac{- ( - 7 ) - 1}{2 \cdot 12} : \frac{1}{4}

\frac{- ( - 7 ) - 1}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{7 - 1}{2 \cdot 12}

Subtract the numbers:

7 - 1 = 6

= \frac{6}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{6}{24}

Cancel the common factor:

6

= \frac{1}{4}

The solutions to the quadratic equation are:

u = \frac{1}{3}, u = \frac{1}{4}

Substitute back

u = cos (x)

cos (x) = \frac{1}{3}, cos (x) = \frac{1}{4}

cos (x) = \frac{1}{3}, cos (x) = \frac{1}{4}

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = \frac{1}{3}

Apply trig inverse properties

cos (x) = \frac{1}{3}

General solutions for

cos (x) = \frac{1}{3}

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

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

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

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

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

Combine all the solutions

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn, x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

Show solutions in decimal form

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

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