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Popular >

cos^2(x)-0.5cos(x)=0

Solution

cos^{2} (x) - 0.5 cos (x) = 0

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) - 0.5 cos (x) = 0

Solve by substitution

cos^{2} (x) - 0.5 cos (x) = 0

Let:

cos (x) = u

u^{2} - 0.5 u = 0

u^{2} - 0.5 u = 0 : u = \frac{1}{2}, u = 0

u^{2} - 0.5 u = 0

Multiply both sides by

10

u^{2} - 0.5 u = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

u^{2} \cdot 10 - 0.5 u \cdot 10 = 0 \cdot 10

Refine

10 u^{2} - 5 u = 0

10 u^{2} - 5 u = 0

Solve with the quadratic formula

10 u^{2} - 5 u = 0

Quadratic Equation Formula:

For

a = 10, b = - 5, c = 0

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

(- 5)^{2} - 4 \cdot 10 \cdot 0 = 5

(- 5)^{2} - 4 \cdot 10 \cdot 0

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 10 \cdot 0

Apply rule

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

Apply radical rule:

assuming

a \geq 0

= 5

u_{1, 2} = \frac{- ( - 5 ) \pm 5}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 5}{2 \cdot 10}, u_{2} = \frac{- ( - 5 ) - 5}{2 \cdot 10}

u = \frac{- ( - 5 ) + 5}{2 \cdot 10} : \frac{1}{2}

\frac{- ( - 5 ) + 5}{2 \cdot 10}

Apply rule

- (- a) = a

= \frac{5 + 5}{2 \cdot 10}

Add the numbers:

5 + 5 = 10

= \frac{10}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{10}{20}

Cancel the common factor:

10

= \frac{1}{2}

u = \frac{- ( - 5 ) - 5}{2 \cdot 10} : 0

\frac{- ( - 5 ) - 5}{2 \cdot 10}

Apply rule

- (- a) = a

= \frac{5 - 5}{2 \cdot 10}

Subtract the numbers:

5 - 5 = 0

= \frac{0}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{0}{20}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 0

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = 0

cos (x) = \frac{1}{2}, cos (x) = 0

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

General solutions for

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

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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