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Popular >

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

Solution

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

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = - 0.84806 \dots + 2 πn, x = π + 0.84806 \dots + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = - 48.59037 \dots^{\circ} + 36 0^{\circ} n, x = 228.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

- \frac{1}{2}

from both sides

sin (x) - 2 cos (2 x) + \frac{1}{2} = 0

Simplify

sin (x) - 2 cos (2 x) + \frac{1}{2} : \frac{2 sin ( x ) - 4 cos ( 2 x ) + 1}{2}

sin (x) - 2 cos (2 x) + \frac{1}{2}

Convert element to fraction:

sin (x) = \frac{sin ( x ) 2}{2}, 2 cos (2 x) = \frac{2 cos ( 2 x ) 2}{2}

= \frac{sin ( x ) \cdot 2}{2} - \frac{2 cos ( 2 x ) \cdot 2}{2} + \frac{1}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( x ) \cdot 2 - 2 cos ( 2 x ) \cdot 2 + 1}{2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 sin ( x ) - 4 cos ( 2 x ) + 1}{2}

\frac{2 sin ( x ) - 4 cos ( 2 x ) + 1}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 sin (x) - 4 cos (2 x) + 1 = 0

Rewrite using trig identities

1 + 2 sin (x) - 4 cos (2 x)

Use the Double Angle identity:

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

= 1 + 2 sin (x) - 4 (1 - 2 sin^{2} (x))

Simplify

1 + 2 sin (x) - 4 (1 - 2 sin^{2} (x)) : 8 sin^{2} (x) + 2 sin (x) - 3

1 + 2 sin (x) - 4 (1 - 2 sin^{2} (x))

Expand

- 4 (1 - 2 sin^{2} (x)) : - 4 + 8 sin^{2} (x)

- 4 (1 - 2 sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = 2 sin^{2} (x)

= - 4 \cdot 1 - (- 4) \cdot 2 sin^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 4 \cdot 1 + 4 \cdot 2 sin^{2} (x)

Simplify

- 4 \cdot 1 + 4 \cdot 2 sin^{2} (x) : - 4 + 8 sin^{2} (x)

- 4 \cdot 1 + 4 \cdot 2 sin^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

= - 4 + 4 \cdot 2 sin^{2} (x)

Multiply the numbers:

4 \cdot 2 = 8

= - 4 + 8 sin^{2} (x)

= - 4 + 8 sin^{2} (x)

= 1 + 2 sin (x) - 4 + 8 sin^{2} (x)

Simplify

1 + 2 sin (x) - 4 + 8 sin^{2} (x) : 8 sin^{2} (x) + 2 sin (x) - 3

1 + 2 sin (x) - 4 + 8 sin^{2} (x)

Group like terms

= 2 sin (x) + 8 sin^{2} (x) + 1 - 4

Add/Subtract the numbers:

1 - 4 = - 3

= 8 sin^{2} (x) + 2 sin (x) - 3

= 8 sin^{2} (x) + 2 sin (x) - 3

= 8 sin^{2} (x) + 2 sin (x) - 3

- 3 + 2 sin (x) + 8 sin^{2} (x) = 0

Solve by substitution

- 3 + 2 sin (x) + 8 sin^{2} (x) = 0

Let:

sin (x) = u

- 3 + 2 u + 8 u^{2} = 0

- 3 + 2 u + 8 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{3}{4}

- 3 + 2 u + 8 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 8, b = 2, c = - 3

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 8 ( - 3 )}{2 \cdot 8}

2^{2} - 4 \cdot 8 (- 3) = 10

2^{2} - 4 \cdot 8 (- 3)

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 8 \cdot 3

Multiply the numbers:

4 \cdot 8 \cdot 3 = 96

= 2^{2} + 96

2^{2} = 4

= 4 + 96

Add the numbers:

4 + 96 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 2 + 10}{2 \cdot 8}, u_{2} = \frac{- 2 - 10}{2 \cdot 8}

u = \frac{- 2 + 10}{2 \cdot 8} : \frac{1}{2}

\frac{- 2 + 10}{2 \cdot 8}

Add/Subtract the numbers:

- 2 + 10 = 8

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8}{16}

Cancel the common factor:

8

= \frac{1}{2}

u = \frac{- 2 - 10}{2 \cdot 8} : - \frac{3}{4}

\frac{- 2 - 10}{2 \cdot 8}

Subtract the numbers:

- 2 - 10 = - 12

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 12}{16}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{12}{16}

Cancel the common factor:

4

= - \frac{3}{4}

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - \frac{3}{4}

Substitute back

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = - \frac{3}{4}

sin (x) = \frac{1}{2}, sin (x) = - \frac{3}{4}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

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

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

sin (x) = - \frac{3}{4}

Apply trig inverse properties

sin (x) = - \frac{3}{4}

General solutions for

sin (x) = - \frac{3}{4}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = arcsin (- \frac{3}{4}) + 2 πn, x = π + arcsin (\frac{3}{4}) + 2 πn

Show solutions in decimal form

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = - 0.84806 \dots + 2 πn, x = π + 0.84806 \dots + 2 πn

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