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

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

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Popular >

2cos^2(x)-7cos(x)=0

Solution

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

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

2 u^{2} - 7 u = 0

2 u^{2} - 7 u = 0 : u = \frac{7}{2}, u = 0

2 u^{2} - 7 u = 0

Solve with the quadratic formula

2 u^{2} - 7 u = 0

Quadratic Equation Formula:

For

a = 2, b = - 7, c = 0

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 0 = 7

(- 7)^{2} - 4 \cdot 2 \cdot 0

Apply exponent rule:

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

n

is even

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

= 7^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 7^{2} - 0

7^{2} - 0 = 7^{2}

= 7^{2}

Apply radical rule:

assuming

a \geq 0

= 7

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

7 + 7 = 14

= \frac{14}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{14}{4}

Cancel the common factor:

2

= \frac{7}{2}

u = \frac{- ( - 7 ) - 7}{2 \cdot 2} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

7 - 7 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = \frac{7}{2}, u = 0

Substitute back

u = cos (x)

cos (x) = \frac{7}{2}, cos (x) = 0

cos (x) = \frac{7}{2}, cos (x) = 0

cos (x) = \frac{7}{2} :

No Solution

cos (x) = \frac{7}{2}

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Popular Examples

2cos^2(x)-7cos(x)=3

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

tan(x)=tan(2x)

tan (x) = tan (2 x)

2cos(x)= 1/2

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

2cos^2(x)-1=1-sin^2(x)

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

cos(x)+0.65=0

cos (x) + 0.65 = 0

Frequently Asked Questions (FAQ)

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