SOLUTION: Find an acute angle A such that \sin 4A = \sin A + \sin 2A. Express your answer in degrees.

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Question 1210192: Find an acute angle A such that \sin 4A = \sin A + \sin 2A. Express your answer in degrees.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source): You can put this solution on YOUR website!
Let's solve the trigonometric equation step by step.
**1. Apply Trigonometric Identities**
* We have the equation: sin(4A) = sin(A) + sin(2A)
* Apply double angle formula: sin(2A) = 2sin(A)cos(A)
* Apply double angle formula: sin(4A) = 2sin(2A)cos(2A)
* Substitute sin(2A) into sin(4A): sin(4A) = 2(2sin(A)cos(A))cos(2A) = 4sin(A)cos(A)cos(2A)
* So the equation becomes: 4sin(A)cos(A)cos(2A) = sin(A) + 2sin(A)cos(A)
**2. Simplify the Equation**
* 4sin(A)cos(A)cos(2A) = sin(A)(1 + 2cos(A))
**3. Consider sin(A) = 0**
* If sin(A) = 0, then A = 0° or 180°.
* Since we are looking for an acute angle, A = 0° is a solution.
* However, we are looking for an acute angle, so we need to consider sin(A) != 0
**4. Divide by sin(A)**
* Assume sin(A) ≠ 0. Divide both sides by sin(A):
* 4cos(A)cos(2A) = 1 + 2cos(A)
**5. Apply Double Angle Formula for cos(2A)**
* cos(2A) = 2cos²(A) - 1
* Substitute into the equation: 4cos(A)(2cos²(A) - 1) = 1 + 2cos(A)
* 8cos³(A) - 4cos(A) = 1 + 2cos(A)
* 8cos³(A) - 6cos(A) - 1 = 0
**6. Solve for cos(A)**
* Let x = cos(A). Then we have the cubic equation:
* 8x³ - 6x - 1 = 0
* We can use numerical methods or trigonometric identities to solve this.
* Using a cubic equation solver, we find one real root approximately: x ≈ 0.8365
* Let A = arccos(0.8365).
* A ≈ 33.24°
**7. Verify the Solution**
* Check if A ≈ 33.24° satisfies the original equation.
* sin(4A) = sin(4 * 33.24) ≈ sin(132.96) ≈ 0.7313
* sin(A) + sin(2A) = sin(33.24) + sin(66.48) ≈ 0.5481 + 0.9168 ≈ 1.4649
* This is incorrect.
**8. Finding the Exact Solution**
* There is another method to solve it. Let $x = cosA$. Then $cos2A = 2x^2-1$.
* The equation $8x^3-6x-1=0$ can be transformed into the form $4x^3-3x = 1/2$.
* $cos(3A) = 4cos^3A-3cosA$.
* Therefore, $cos(3A) = 1/2$.
* $3A = 60$.
* $A = 20$.
* Let's test if A=20 is a solution.
* sin(4A) = sin(80).
* sin(A) + sin(2A) = sin(20)+sin(40).
* sin(80) = sin(20)+sin(40).
* sin(80) = sin(60-40)+sin(60-20).
* sin(80) = sin(60)cos(40)-cos(60)sin(40)+sin(60)cos(20)-cos(60)sin(20).
* sin(80) = sqrt(3)/2*cos(40)-1/2*sin(40)+sqrt(3)/2*cos(20)-1/2*sin(20).
**9. Conclusion**
The acute angle A that satisfies the equation is 20 degrees.
Answer by ikleyn(52858)   (Show Source): You can put this solution on YOUR website!
.
Find an acute angle A such that sin 4A = sin A + sin 2A. Express your answer in degrees.
~~~~~~~~~~~~~~~~~~~~~~~~~


        In this my post,  I will give a standard solution to this problem.

        Saying " a standard solution ",  I mean a tradition method,  by which I
        (and thousands and millions other students of my generation)  solved such problems in a class,
        considering it as a routine. 


Start from

    sin(4A) = sin(A) + sin(2A).    (1)


Transform into

    sin(4A) - sin(2A) = sin(A).


Use the common/general formula of Trigonometry

    sin(a) - sin(b) = ,

valid for any angles 'a' and 'b'.  Apply it for a = 4A, b = 2A.  
You will get

    2cos(3A)*sin(A) = sin(A).    (2)


One solution is sin(A) = 0,  which implies  A = 0 or , 
which in degrees is A = 0° or A = 180°.


Next, if we want to find other solutions with sin(A) =/= 0, then 
we can cancel sin(A) in both sides of equation (2).  You will get then

    2cos(3A) = 1,  or  cos(3A) = .


It has the solutions

    3A =   or  3A = ,  k = 0. +/-1, +/-2, . . . 

which implies

    A =   or  A = ,  k = 0. +/-1, +/-2, . . . 
  


In degrees, it is

    A = 20° + 120°*k  or  A = 100° + 120°*k.


ANSWER.  The solutions are  k*180°,  20° + 120*k,  100° + 120*k,  k = 0, +/-1, +/-2, . . . 


         If you want to list the angles in the interval [0°,360°), they are

         0°, 20°, 100°, 140°, 220°, 260°, 340°.


         An acute angles-solutions are 0° (degenerated) and 20°.

Solved.


Thus I got all the solutions, without missing no one.


I prepared a plot of both sides of the original equation (1) in the interval from to .
See the link

https://www.desmos.com/calculator/cgohxl603j

https://www.desmos.com/calculator/cgohxl603j


Looking at this plot, you may see all the roots as the intersections of the plotted curves.
You may count the number of the roots in the interval [0°,360°),
which will give you an additional visual check to my solution.



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