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5.5.Trigonometric Equations

Thus far in this chapter we have derived a host of useful trigonometric formulas. In this chapter, we will use these formulas to solve various trigonometric equations of one variable. These are equations of the form f(θ) = A, where f(θ) is a trigonometric function of the variable θ and A is a constant. The difficulty of solving such equations depends on how complicated f is.

Note that if θ is an angle, as we assume it is throughout this section, then it is only necessary to solve for θ in the interval [0, 2π), since adding any integer multiple of 2π will yield a new solution.

**Example 1: **Solve the trigonometric equation sin θ = 0.

**Solution: **The obvious solution, which from now on we call the basic solution, is θ = arcsin 0 = 0. But from Equation (4.4.3), namely the identity sin(π - θ) = sin θ, we see that whenever θ = θ_{0} is a solution to an equation of the form sin θ = A , then θ = π - θ_{0} is as well. Since in this case θ_{0} = 0 is a solution, we see that θ = π - θ_{0} = π is the other one. (There are no more solutions, since the sine function attains every value at most twice per period.

**Example 2: ** Solve the trigonometric equation sin θ = 1/2.

**Solution: **The basic solution is θ = arcsin 1/2 = π/6. By virtue of the remarks in the solution to the previous example, we see that π - π/6 = 5π/6 is the other one.

**Example 3: ** Solve the trigonometric equation cos θ = -1/2.

**Solution: **The basic solution is θ = arccos(-1/2) = 2π/3. Since the cosine function is even, we know that θ = -2π/3 is another solution. Since this does not lie in the interval [0, 2π), we add the appropriate multiple of 2π to it to make it lie in the interval. 2π does the job, yielding the second solution θ = -2π/3 + 2π = 4π/3.

**Example 4: ** Solve the trigonometric equation tan θ = √3.

**Solution: **The basic solution is θ = arctan(√3) = π/3. But we know the tangent function is periodic with period π and monotonic within every period, whence it attains every value exactly once within a period. Thus, the only other solution is θ = π/3 + π = 4π/3.

**Example 5: ** Solve the trigonometric equation sin 2θ = √3/2.

**Solution: **It is easiest here to solve for 2θ first in the interval [0,4π). Our basic solution is 2θ = arcsin(√3/2) = π/3. From the remarks following Examples 1 and 2, we see that 2θ = π - π/3 = 2π/3 is another solutions. The two other solutions in the interval [0,4π) are obtained from these by adding 2π; thus 2θ = π/3 + 2π = 7π/3 and 2θ = 2π/3 + 2π = 8π/3 are the other two solutions. Dividing each these four solutions by 2, we see that the solutions to the equation are θ = π/6, π/3, 7π/6 and 4π/3.

**Example 6: ** Solve the trigonometric equation sin^{2}θ = 1/2.

**Solution: **Taking the square root of both sides yields sin θ = ±1/√2 = ±√2/2. The two basic solutions are thus θ = arcsin(±√2/2) = ±π/4. To ensure that the solution -π/4 lies in the interval [0,2π), we add 2π to it, yielding θ = 7π/4. Thus we have found the two solutions θ = π/4 and θ = 7π/4. Add to these the solutions θ = π - π/4 = 3π/4 and θ = π - (-π/4) = 5π/4, and we have them all. Thus, the four solutions to this equation are θ = π/4, 3π/4, 5π/4 and 7π/4.

**Example 7: ** Solve the trigonometric equation 2 cos^{4}θ - 3 cos^{2}θ + 1 = 0.

**Solution: **This is a classic quadratic equation in x = cos^{2}θ, which by this substitution becomes f(x) = 2x^{2} - 3x + 1 = 0. By trial and error, we see that f(x) factors as (x - 1)(2x - 1), whence the solution in x are x = 1 and x = 1/2. Recalling that x = cos^{2}θ and solving for cos^{ }θ, we see that the solutions satisfy cos^{ }θ = ±1 or cos^{ }θ = ±1/√2 = ±√2/2. We solve each of these four equations in turn. The equation cos^{ }θ = 1 yields the unique solution θ = 0, while the equation cos^{ }θ = -1 yields the unique solution θ = π. The equation cos θ = √2/2 yields the basic solution θ = arccos(√2/2) = π/4. By virtue of our earlier remarks, we see that θ = 2π - π/4 = 7π/4 is the other solution to this equation. Finally, the equation cos θ = -√2/2 yields the basic solution θ = arccos(-√2/2) = 3π/4, while the remaining solution is 2π - 3π/4 = 5π/4. Thus, we have found the following six solutions to the original equation: θ = 0, π/4, 3π/4, π, 5π/4, and 7π/4.

**Example 8: ** Solve the trigonometric equation cos 2θ - cos θ + 1 = 0.

**Solution: **To solve this equation, we make use of the first of the double-angle formulas in (5.3.3) namely cos 2θ = 2 cos^{2}θ - 1. Upon making this substitution, our equation becomes 2 cos^{2}θ - cos θ = 0, whose solutions satisfy cos θ = 0 or cos θ = 1/2. The equation cos θ = 0 yields the solutions θ = π/2 and θ = 3π/2, while cos θ = 1/2 yields θ = π/3 and θ = 5π/3. Thus, we have found the four solutions θ = π/3, π/2, 3π/2, and 5π/3.

**Example 9: ** Solve the trigonometric equation sin θ + cos θ = 1.

**Solution:** Note that sin θ + cos θ = √2 (sin θ cos π/4 + cos θ sin π/4) = √2 sin(θ + π/4), by virtue of the sum formula (5.2.1) for sine. Thus we have sin(θ + π/4) = 1/√2 = √2/2, whose basic solution is θ + π/4 = arcsin √2/2 = π/4, yielding θ = 0. The other solution is θ + π/4 = π - π/4 = 3π/4, yielding θ = π/2.