9514 1404 393
Answer:
A = 22.5°, 45°, 112.5°, 202.5°, 225°, 292.5°
Explanation:
Use the angle reduction formula to reduce this to an expression involving only sine or cosine, then solve the resulting higher-degree polynomial.
sin(3x) = 3sin(x) -4sin³(x)
Using the identity for cosine, we have the radical equation ...
√(1 -sin²(A)) = 3sin(A) -4sin³(A)
Squaring both sides, we have ...
1 - sin²(A) = 9sin²(A) -24 sin⁴(A) +16sin⁶(A)
Letting z = sin²(A), this is the cubic equation ...
1 -z = 9z -24z² +16z³
16z³ -24z² +10z -1 = 0
Solving this by the usual methods, we find its factors to be ...
(2z -1)(8z² -8z +1) = 0
8(2z -1)((z -1/2)² -1/8) = 0 . . . . quadratic in vertex form
Then the roots are ...
z = (1 ±√(1/2))/2, z = 1/2
The angle solutions are ...
A = arcsin(±√z)
Of course, these need to be adjusted to fall in the domain of interest, and checked for extraneous solutions.
For z = 1/2-√(1/8), the angles are ...
A = ±22.5° ⇒ 22.5°, 157.5°, 202.5°, 337.5°
For z = 1/2, the angles are ...
A = ±45° ⇒ 45°, 135°, 225°, 315°
For z = 1/2+√(1/8), the angles are ...
A = ±67.5° ⇒ 67.5°, 112.5°, 247.5°, 292.5°
The angles that are in bold are the actual solutions; the others are extraneous.
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All of the above is "the hard way." A much simpler approach is to take advantage of the identity ...
cos(a) -sin(b) = (1/2)·sin(b/2 -a/2 -π/4)·sin(b/2 +a/2 -π/4)
Then you can use the zero product rule to find (for a=A, b=3A) ...
(3A/2 -A/2 -π/4) = nπ ⇒ A = nπ +π/4 ⇒ 45°, 225°
(3A/2 +A/2 -π/4) = nπ ⇒ A = (nπ +π/4) ⇒ 22.5°, 112.5°, 202.5°, 292.5°
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Rewriting the equation as a function equal to zero, the solutions are the x-intercepts of the graph. In the attached, the x-axis is set to degrees.
f(A) = cos(A) -sin(3A) = 0