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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