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Difficulty: Very hardTrigonometric Graphs and Simple Equations

Find the smallest positive value of θ\theta (in degrees) satisfying the trigonometric equation sin⁡(3θ)+3cos⁡(3θ)=2\sin(3\theta) + \sqrt{3}\cos(3\theta) = \sqrt{2}.

Answer: 25 degrees

Answer

The smallest positive value of θ\theta is 25∘25^\circ.
Dividing the equation by 22 reduces it to sin⁡(3θ+60∘)=22\sin(3\theta + 60^\circ) = \frac{\sqrt{2}}{2}. The principal acute angle is 45∘45^\circ. The second quadrant angle giving a positive sine is 180∘−45∘=135∘180^\circ - 45^\circ = 135^\circ. Setting 3θ+60∘=135∘3\theta + 60^\circ = 135^\circ gives 3θ=75∘3\theta = 75^\circ, which yields θ=25∘\theta = 25^\circ. This is smaller than any positive solution generated by the first quadrant branch.

Step-by-Step Solution

1
Transform the left-hand side into a single harmonic function Rsin⁡(3θ+α)R\sin(3\theta + \alpha).
Dividing the equation by 22 yields 12sin⁡(3θ)+32cos⁡(3θ)=22\frac{1}{2}\sin(3\theta) + \frac{\sqrt{3}}{2}\cos(3\theta) = \frac{\sqrt{2}}{2}, which simplifies to sin⁡(3θ+60∘)=22\sin(3\theta + 60^\circ) = \frac{\sqrt{2}}{2}.
The identity sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A + B) = \sin A \cos B + \cos A \sin B allows us to combine sin⁡(3θ)\sin(3\theta) and cos⁡(3θ)\cos(3\theta) using cos⁡60∘=12\cos 60^\circ = \frac{1}{2} and sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}.
2
Find the quadrant solutions for the angle 3θ+60∘3\theta + 60^\circ.
First quadrant: 3θ+60∘=45∘+360∘k  ⟹  3θ=−15∘+360∘k3\theta + 60^\circ = 45^\circ + 360^\circ k \implies 3\theta = -15^\circ + 360^\circ k.
Second quadrant: 3θ+60∘=135∘+360∘k  ⟹  3θ=75∘+360∘k3\theta + 60^\circ = 135^\circ + 360^\circ k \implies 3\theta = 75^\circ + 360^\circ k.
Since the sine of the angle is positive (22\frac{\sqrt{2}}{2}), solutions exist in both the first (45∘45^\circ) and second (180∘−45∘=135∘180^\circ - 45^\circ = 135^\circ) quadrants.
3
Calculate the smallest positive angle θ\theta.
For k=0k = 0 in the second quadrant branch, 3θ=75∘  ⟹  θ=25∘3\theta = 75^\circ \implies \theta = 25^\circ. (The first quadrant branch yields θ=−5∘\theta = -5^\circ for k=0k=0 and θ=115∘\theta = 115^\circ for k=1k=1). Thus, θ=25∘\theta = 25^\circ is the smallest positive solution.
Comparing all non-negative resulting angles demonstrates that 25∘25^\circ is the smallest strictly positive solution.

Key Concept

Solving linear trigonometric equations of the form Asin⁡x+Bcos⁡x=CA\sin x + B\cos x = C using RR-formula reduction.
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Topics

Trigonometric Graphs and Simple Equations

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