Solve the following equation for all solutions: sin(11x)cos(6x)-cos(11x)sin(6x)=√3/2
sin ( A - B ) = sin A ∙ cos B - cos A ∙ sin B
that's why
sin ( 11 x ) ∙ cos ( 6 x ) - cos ( 11 x ) ∙ sin( 6 x ) = sin ( 11 x - 6 x ) = sin 5 x = √3 / 2
sin 5 x = √3 / 2
5 x = sin⁻¹ ( √3 / 2 ) = arcsin ( √3 / 2 )
The solutions of this equuation are:
5 x = π / 3 and 5 x = 2 π / 3
Period of sin x = 2 π , so the solutions of equuation sin 5 x = √3 / 2 are:
5 x = π / 3 + 2 π n and 5 x = 2 π / 3 + 2 π n
where:
n = ± 1 , ± 2 , ± 3...
Divide both sides by 5
x = π / 15 + 2 π n / 5 and x = 2 π / 15 + 2 π n / 5
sin(11x)cos(6x)-cos(11x)sin(6x)=√3/2
did you notice the pattern on the left?
sin(A-B) = sinAcosB - cosAsinB
so
sin(11x)cos(6x)-cos(11x)sin(6x)=√3/2
sin(11x - 6x) = √3/2
sin 5x = √3/2
5x = 60° or 120°
x = 15 or 24°
the period of sin 5x = 360/5° = 72°
you asked for all solutions, which is not possible, since we have an infinite number of solutions,
the general solution would be
x = 15° + 72k° or 24° + 72k°, where k is an integer, eg. 240° would be a solution
in case you want radians:
x = π/12 + (2π/5)k or 2π/15 ° + (2π/5)k °
To solve the equation sin(11x)cos(6x) - cos(11x)sin(6x) = √3/2, we can use the trigonometric identity for the sine of the difference of two angles:
sin(A - B) = sin(A)cos(B) - cos(A)sin(B)
Comparing this to the equation given, we see that A = 11x and B = 6x. Therefore, we can rewrite the equation as:
sin(11x - 6x) = √3/2
Simplifying the expression inside the sine function:
sin(5x) = √3/2
To find the solutions for this equation, we can use the reference angles of sin inverse (√3/2). The reference angle is π/3 or 60 degrees, which means that sin(θ) = √3/2 when θ = π/3 or θ = 2π/3.
Now we can solve for x:
5x = π/3 or 5x = 2π/3
Dividing both sides by 5:
x = π/15 or x = 2π/15
To find all solutions, we can add multiples of the period of the sine function, which is 2π/5, to both sides of the equation.
For the first solution, x = π/15:
x = π/15 + 2π/5 = (π + 6π)/15 = 7π/15
For the second solution, x = 2π/15:
x = 2π/15 + 2π/5 = (2π + 6π)/15 = 8π/15
Therefore, the solutions to the equation sin(11x)cos(6x) - cos(11x)sin(6x) = √3/2 are x = 7π/15 and x = 8π/15.
To solve the equation sin(11x)cos(6x)-cos(11x)sin(6x) = √3/2, we can start by using the angle addition formula for sine:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
Let's rewrite the equation using the angle addition formula:
sin(11x + 6x) = √3/2
Simplifying further:
sin(17x) = √3/2
To find the solutions, we need to determine when the sine function takes on the value of √3/2. This occurs at certain angles in the unit circle, where the y-coordinate is √3/2.
The values of sin(θ) = √3/2 can be found for the angles θ = π/3 and θ = 2π/3 (among others). These angles correspond to positions of the terminal side of an angle in the unit circle.
Now, we need to find the values of x that satisfy sin(17x) = √3/2. Since the sine function has a period of 2π, the solutions will occur when 17x is equal to π/3 plus any integer multiple of 2π or when 17x is equal to 2π/3 plus any integer multiple of 2π.
Let's set up the equations:
17x = π/3 + 2πn where n is an integer
17x = 2π/3 + 2πm where m is an integer
Solving these equations for x, we get:
x = (π/3 + 2πn)/17 where n is an integer
x = (2π/3 + 2πm)/17 where m is an integer
These equations give us the possible solutions for x. To find all solutions, we can substitute different integer values for n and m and calculate the corresponding x values.
Note: Since the equation involves trigonometric functions, there may be an infinite number of solutions, depending on the range of x we are considering.