What is the general solution for 4cos^2(x)-7cos(x)+3=0 ?

The general solution for 4cos^2(x)-7cos(x)+3=0 is x=2pin,x=0.72273…+2pin,x=2pi-0.72273…+2pin

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Popular >

4cos^2(x)-7cos(x)+3=0

Solution

4 cos^{2} (x) - 7 cos (x) + 3 = 0

Solution

x = 2 πn, x = 0.72273 \dots + 2 πn, x = 2 π - 0.72273 \dots + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 41.40962 \dots^{\circ} + 36 0^{\circ} n, x = 318.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos^{2} (x) - 7 cos (x) + 3 = 0

Solve by substitution

4 cos^{2} (x) - 7 cos (x) + 3 = 0

Let:

cos (x) = u

4 u^{2} - 7 u + 3 = 0

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

4 u^{2} - 7 u + 3 = 0

Solve with the quadratic formula

4 u^{2} - 7 u + 3 = 0

Quadratic Equation Formula:

For

a = 4, b = - 7, c = 3

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

(- 7)^{2} - 4 \cdot 4 \cdot 3 = 1

(- 7)^{2} - 4 \cdot 4 \cdot 3

Apply exponent rule:

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

n

is even

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

= 7^{2} - 4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 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 4}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 4}, u_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 4}

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

\frac{- ( - 7 ) + 1}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{7 + 1}{2 \cdot 4}

Add the numbers:

7 + 1 = 8

= \frac{8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 7 ) - 1}{2 \cdot 4} : \frac{3}{4}

\frac{- ( - 7 ) - 1}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{7 - 1}{2 \cdot 4}

Subtract the numbers:

7 - 1 = 6

= \frac{6}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{6}{8}

Cancel the common factor:

2

= \frac{3}{4}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

General solutions for

cos (x) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = 2 πn, x = 0.72273 \dots + 2 πn, x = 2 π - 0.72273 \dots + 2 πn

Popular Examples

tan(θ)=(-4/3-(-11/3))/(1+(-4/3)(-11/3))

tan (θ) = \frac{- \frac{4}{3} - ( - \frac{11}{3} )}{1 + ( - \frac{4}{3} ) ( - \frac{11}{3} )}

8+2cot(x)=12

8 + 2 cot (x) = 12

(1+cot(p))tan(p)=3

(1 + cot (p)) tan (p) = 3

csc(θ+pi)=-(2sqrt(3))/3

csc (θ + π) = - \frac{2 3}{3}

sin(x)= 33/105

sin (x) = \frac{33}{105}

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-7cos(x)+3=0 ?
The general solution for 4cos^2(x)-7cos(x)+3=0 is x=2pin,x=0.72273…+2pin,x=2pi-0.72273…+2pin