What is the general solution for cos(2x)cos(x)-sin(2x)sin(x)=0.5 ?

The general solution for cos(2x)cos(x)-sin(2x)sin(x)=0.5 is x= pi/9+(2pin)/3 ,x=(5pi)/9+(2pin)/3

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cos(2x)cos(x)-sin(2x)sin(x)=0.5

Solution

cos (2 x) cos (x) - sin (2 x) sin (x) = 0.5

Solution

x = \frac{π}{9} + \frac{2 πn}{3}, x = \frac{5 π}{9} + \frac{2 πn}{3}

Degrees

x = 2 0^{\circ} + 12 0^{\circ} n, x = 10 0^{\circ} + 12 0^{\circ} n

Solution steps

cos (2 x) cos (x) - sin (2 x) sin (x) = 0.5

Rewrite using trig identities

cos (2 x) cos (x) - sin (2 x) sin (x)

Use the Angle Sum identity:

cos (s) cos (t) - sin (s) sin (t) = cos (s + t)

= cos (2 x + x)

cos (2 x + x) = 0.5

General solutions for

cos (2 x + x) = 0.5

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}

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

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

Solve

2 x + x = \frac{π}{3} + 2 πn : x = \frac{π}{9} + \frac{2 πn}{3}

2 x + x = \frac{π}{3} + 2 πn

Add similar elements:

2 x + x = 3 x

3 x = \frac{π}{3} + 2 πn

Divide both sides by

3

3 x = \frac{π}{3} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{π}{3}}{3} + \frac{2 πn}{3} : \frac{π}{9} + \frac{2 πn}{3}

\frac{\frac{π}{3}}{3} + \frac{2 πn}{3}

\frac{\frac{π}{3}}{3} = \frac{π}{9}

\frac{\frac{π}{3}}{3}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{π}{9}

= \frac{π}{9} + \frac{2 πn}{3}

x = \frac{π}{9} + \frac{2 πn}{3}

x = \frac{π}{9} + \frac{2 πn}{3}

x = \frac{π}{9} + \frac{2 πn}{3}

Solve

2 x + x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{9} + \frac{2 πn}{3}

2 x + x = \frac{5 π}{3} + 2 πn

Add similar elements:

2 x + x = 3 x

3 x = \frac{5 π}{3} + 2 πn

Divide both sides by

3

3 x = \frac{5 π}{3} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3} : \frac{5 π}{9} + \frac{2 πn}{3}

\frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3}

\frac{\frac{5 π}{3}}{3} = \frac{5 π}{9}

\frac{\frac{5 π}{3}}{3}

Apply the fraction rule:

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

= \frac{5 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{5 π}{9}

= \frac{5 π}{9} + \frac{2 πn}{3}

x = \frac{5 π}{9} + \frac{2 πn}{3}

x = \frac{5 π}{9} + \frac{2 πn}{3}

x = \frac{5 π}{9} + \frac{2 πn}{3}

x = \frac{π}{9} + \frac{2 πn}{3}, x = \frac{5 π}{9} + \frac{2 πn}{3}

Popular Examples

sin^2(θ)=0.4567

sin^{2} (θ) = 0.4567

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

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

cos(θ)= 10/15

cos (θ) = \frac{10}{15}

sin(θ)= 40/60

sin (θ) = \frac{40}{60}

10=3.44sin(0.51t+4.11)+9.48

10 = 3.44 sin (0.51 t + 4.11) + 9.48

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)cos(x)-sin(2x)sin(x)=0.5 ?
The general solution for cos(2x)cos(x)-sin(2x)sin(x)=0.5 is x= pi/9+(2pin)/3 ,x=(5pi)/9+(2pin)/3