Solve trig equations and find all solutions in a given interval. Math 30-1 Alberta.
Lesson 5.7 of Trigonometric Identities in Math 30-1 — Alberta curriculum lessons.
To solve a trigonometric equation means to determine all possible angle values that make the equation true within a given domain. When the domain is not restricted, we need to provide a general solution.
When a trigonometric equation involves more than one trig ratio, it is sometimes helpful to use identities to rewrite the equation in terms of a single ratio by using one or more of the following:
Once the equation is rewritten in terms of a single trig ratio and set equal to zero, use algebra to simplify and determine the solutions.
📌 Remember. Always check that your solutions do not include non-permissible values from the original equation. We can always verify solutions graphically.
Part a) 2x + x = 0. Step 1 — Check for NPVs: None for this equation — and are defined for all real numbers, so no values need to be excluded.
Step 2 — Identify the problem: We have two different arguments: 2x and x. We cannot solve an equation that mixes 2x and x directly — they refer to different angles. Our goal is to rewrite everything in terms of a single trig ratio with the same argument.
Step 3 — Choose and apply the identity: The Double Angle Identity tells us:
2x = 2 x x
We use this because 2x is exactly double the argument x, so this identity converts 2x into an expression involving only x and x. Substituting:
2 x x + x = 0
Step 4 — Factor: Both terms share a common factor of x. Factor it out:
x(2 x + 1) = 0
This is the crucial algebra step — by factoring, we split one hard equation into two simple ones.
Step 5 — Apply the Zero Product Property: If a product equals zero, at least one factor must be zero:
x = 0 or 2 x + 1 = 0 → x = -(1)/(2)
Step 6 — Solve each equation over [0, 2π):
x = 0 at x = 0 and x = π (where the sine curve crosses the x-axis)
x = -(1)/(2) at x = (2π)/(3) (Quadrant II) and x = (4π)/(3) (Quadrant III)
The reference angle is (π)/(3) since (π)/(3) = (1)/(2). Cosine is negative in Q2 and Q3.
x = 0,\ (2π)/(3),\ π,\ (4π)/(3)
Part b) ^2 x = x x. Step 1 — Check for NPVs: The term x = ( x)/( x) requires x ≠ 0, so we must exclude:
x ≠ (π)/(2),\ (3π)/(2)
These are the values where x = 0, making x undefined. They are eliminated before we even begin solving.
Step 2 — Identify the problem: The right side contains x, while the left side contains ^2 x. We want a single trig ratio — ideally just x — so we can solve easily.
Step 3 — Choose and apply the identity: We use the Quotient Identity:
x = ( x)/( x)
Substitute into the right side:
^2 x = ( x)/( x) · x
The x appears in the denominator of the fraction AND as a multiplier — they cancel:
^2 x = x x · x
^2 x = x
Now the entire equation is in terms of x only — exactly what we wanted.
Step 4 — Rearrange and factor: Move all terms to one side:
^2 x - x = 0
Factor out the common x:
x( x - 1) = 0
Step 5 — Apply the Zero Product Property: If a product equals zero, at least one factor must be zero. Set each factor equal to zero separately:
x = 0 or x - 1 = 0 → x = 1
Step 6 — Solve each equation over [0, 2π):
x = 0 at x = 0 and x = π
x = 1 at x = (π)/(2)
Step 7 — Check against NPVs: x = (π)/(2) was flagged as an NPV because it makes x = 0 and x undefined. Even though it appeared as a solution, it must be rejected.
x = 0,\ π
Part c) ^2 x - 3^2 x = 0. Step 1 — Check for NPVs: None — and are defined for all real numbers.
Step 2 — Identify the problem: We have two different trig ratios: ^2 x and ^2 x. We cannot directly solve for x when two different ratios are mixed. The strategy is to eliminate one of them by converting to a single ratio.
Step 3 — Choose and apply the identity: From the Pythagorean Identity:
^2 x + ^2 x = 1
Rearranging gives us:
^2 x = 1 - ^2 x
We substitute (1 - ^2 x) in place of ^2 x. The ^2 x disappears — replaced entirely by terms:
^2 x_1 - ^2 x - 3^2 x = 0
1 - ^2 x - 3^2 x = 0
Step 4 — Simplify: Combine like terms:
1 - 4^2 x = 0
4^2 x = 1
^2 x = (1)/(4)
x = ±(1)/(2)
We take both the positive and negative square root because squaring loses sign information.
Step 5 — Solve for x over [0, 2π):
The reference angle is (π)/(6) since (π)/(6) = (1)/(2).
x = +(1)/(2): positive in Q1 and Q2 → x = (π)/(6), (5π)/(6)
x = -(1)/(2): negative in Q3 and Q4 → x = (7π)/(6), (11π)/(6)
x = (π)/(6),\ (5π)/(6),\ (7π)/(6),\ (11π)/(6)
Part d) 2x - 2 x + 3 = 0. Step 1 — Check for NPVs: None — and are defined for all real numbers.
Step 2 — Identify the problem: Similar to Part a), we have a mixed argument: 2x (argument 2x) and x (argument x). We need to unify the argument. We also notice there is a x term already present, so our goal is to rewrite 2x using x.
Step 3 — Choose the right form of the identity: The Double Angle Identity has three equivalent forms:
2x = ^2 x - ^2 x = 2^2 x - 1 = 1 - 2^2 x
We choose 2x = 1 - 2^2 x because it converts 2x entirely into x — matching the other term already in the equation. Using a different form would leave a term behind, making things harder.
Substituting:
(1 - 2^2 x) - 2 x + 3 = 0
-2^2 x - 2 x + 4 = 0
Step 4 — Simplify and factor: Factor out -2:
-2(^2 x + x - 2) = 0
Divide both sides by -2:
^2 x + x - 2 = 0
This is a quadratic in x — treat x like a variable (let u = x):
(u - 1)(u + 2) = 0
x = 1 or x = -2
Step 5 — Eliminate impossible values: The range of sine is [-1, 1], so x = -2 has no solution. Reject it.
x = 1 occurs at x = (π)/(2) within [0, 2π).
x = (π)/(2)
📌 Key Takeaway from Example 1. Parts a) and d) both involved a change of argument (from 2x to x) using the Double Angle Identity. The key decision was which form of 2x to use — always pick the form that matches the other ratio already in the equation. Parts b) and c) involved eliminating a mixed ratio using the Quotient and Pythagorean identities respectively.
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