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MathHardAdv 2010 • Paper 1

Number of Theta Values for Non-trivial Solution of Trigonometric System

The number of all possible values of θ\theta, where 0<θ<π0 < \theta < \pi, for which the system of equations

(y+z)cos⁡3θ=(xyz)sin⁡3θxsin⁡3θ=2cos⁡3θy+2sin⁡3θz(xyz)sin⁡3θ=(y+2z)cos⁡3θ+ysin⁡3θ\begin{array}{c} (y + z) \cos 3\theta = (xyz) \sin 3\theta \\ x \sin 3\theta = \frac{2 \cos 3\theta}{y} + \frac{2 \sin 3\theta}{z} \\ (xyz) \sin 3\theta = (y + 2z) \cos 3\theta + y \sin 3\theta \end{array}

have a solution (x0,y0,z0)(x_0, y_0, z_0) with y0z0≠0y_0 z_0 \neq 0, is

Official Numerical Answer3

Topics & Concepts

Trigonometric FunctionsTrigonometric Equations

Step-by-Step Solution

To find the number of all possible values of θ∈(0,π)\theta \in (0, \pi) for which the given system of equations has a solution (x0,y0,z0)(x_0, y_0, z_0) with y0z0≠0y_0 z_0 \neq 0, let us simplify the system.

Let C=cos⁡3θC = \cos 3\theta and S=sin⁡3θS = \sin 3\theta. The given system of equations is:

(y+z)C=xyzS— (1)(y + z) C = xyz S \quad \text{--- (1)} xS=2Cy+2Sz— (2)x S = \frac{2 C}{y} + \frac{2 S}{z} \quad \text{--- (2)} xyzS=(y+2z)C+yS— (3)xyz S = (y + 2z) C + y S \quad \text{--- (3)}

Since y0z0≠0y_0 z_0 \neq 0, we can multiply equation (2) by yzyz to get:

xyzS=2zC+2yS— (2’)xyz S = 2 z C + 2 y S \quad \text{--- (2')}

Equating the right-hand sides of (1) and (3):

(y+z)C=(y+2z)C+yS(y + z) C = (y + 2z) C + y S yC+zC=yC+2zC+ySy C + z C = y C + 2z C + y S zC=2zC+yS  ⟹  yS+zC=0  ⟹  yS=−zC— (4)z C = 2z C + y S \implies y S + z C = 0 \implies y S = -z C \quad \text{--- (4)}

Substitute equation (4) into equation (2'):

xyzS=2zC+2(yS)=2zC+2(−zC)=0xyz S = 2 z C + 2 (y S) = 2 z C + 2 (-z C) = 0

Thus, from equation (1), we have:

(y+z)C=xyzS=0(y + z) C = xyz S = 0

This gives two cases to consider: C=0C = 0 or y+z=0y + z = 0.

Case 1: If C=0C = 0
If cos⁡3θ=0\cos 3\theta = 0, then S=sin⁡3θ=±1≠0S = \sin 3\theta = \pm 1 \neq 0.
From equation (4), yS=−z(0)=0  ⟹  y=0y S = -z(0) = 0 \implies y = 0 (since S≠0S \neq 0).
However, this contradicts the given condition y0z0≠0y_0 z_0 \neq 0. Thus, C≠0C \neq 0.

Case 2: If y+z=0y + z = 0
Since C≠0C \neq 0, we must have:

y+z=0  ⟹  z=−yy + z = 0 \implies z = -y

Substitute z=−yz = -y into equation (4):

yS=−(−y)C=yCy S = -(-y) C = y C

Since y0≠0y_0 \neq 0, we can divide by yy:

S=C  ⟹  sin⁡3θ=cos⁡3θ  ⟹  tan⁡3θ=1S = C \implies \sin 3\theta = \cos 3\theta \implies \tan 3\theta = 1

For tan⁡3θ=1\tan 3\theta = 1, we have S=C≠0S = C \neq 0, x0=0x_0 = 0, and z0=−y0≠0z_0 = -y_0 \neq 0, which consistently satisfies all three original equations.

Given 0<θ<π0 < \theta < \pi, the range for 3θ3\theta is:

0<3θ<3π0 < 3\theta < 3\pi

Solving tan⁡3θ=1\tan 3\theta = 1 in this domain:

3θ=π4,5π4,9π43\theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}

Dividing by 3 gives the values of θ\theta:

θ=π12,5π12,3π4\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4}

All three values lie within the interval (0,π)(0, \pi).

Thus, the number of all possible values of θ\theta is 33.

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