5.3 (Solving Trigonometric Equations)

- Introduction:
To solve a trigonometric equation, use standard algebraic techniques such as:

- collecting like terms
- factoring

The main goal is to isolate the trigonometric function involved in the equation.
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- Solving a trigonometric equation:

- First, isolate the trigonometric function, in this case, sinx:

2 sinx - 1 = 0 original equation

2 sinx = 1 add 1 to each side.
sinx = 1/2 divide each side by 2.

- Second, solve for x:

sinx = 1/2, therefore, it has the solutions x = π/6 and x = 5π/6

*NOTE: most of the time solutions will be within the interval [0, 2π), however, any angle that is coterminal with the solution found from the equation can be a solution as well.
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- Collecting like terms:
- First, rewrite the equation so that the trigonometric function, sinx, is isolated on one side of the equation.

sinx + √2 = -sinx write original equation.

sinx + sinx = -√2 add sinx and subtract √2 from each side.
2 sinx = -√2 combine like terms.
sinx = -√2/2 divide each side by 2.

- Second, find the solutions to x within the interval [0, 2π):

x = 5π/4 and x = 7π/4
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- Extracting Square Roots:
- First, rewrite the equation so that the trigonometric function, tanx, is isolated on one side of the equation.

3 tan²x - 1 = 0 write original equation.
3 tan²x = 1 add 1 to each side.
tan²x = 1/3 divide each side by 3.
tanx = ± 1/√3 extract square roots.

*NOTE: when taking the square root, remember it's ±!
- Second, solve for x:

The solutions are then x = π/6 and x = 5π/6

*NOTE: remember that tanx and cotx have a period of π, therefore, the interval changes to [0, π)
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- Factoring:
- First, solve the equation:

cotx cos²x = 2 cotx write original equation.
cotx cos²x - 2 cotx = 0 subtract 2 cotx from each side.
cotx(cos²x - 2) = 0

*NOTE: do NOT make the mistake of dividing each side of the equation by cotx because solutions will be lost.
- Second, set each factor equal to zero:

cotx = 0
x = π/2

cos²x - 2 = 0
cos²x = 2
cosx = ± √2

*NOTE: check the solutions, notice that ± √2 is not a solution because it is outside of the range of the cosine function.
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- Equations of quadratic type:
- Many trigonometric equations are of the quadratic type.

Quadratic in sinx:
2 sin²x - sinx - 1 = 0
2(sinx)² - sinx - 1 = 0

Quadratic in secx:
sec²x - 3 secx - 2 = 0
(secx)² - 3 secx - 2 = 0

- To solve equations of this type, factor the quadratic or, if factoring is not possible, use the quadratic formula:

(trigonometric function)x = -b ± √b² - 4ac/2a

*NOTE: don't forget the trigonometric function in front of the x!
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- Factoring an equation of quadratic type:
- First, write the original equation and factor:

2 sin²x - sinx - 1 = 0 write the original equation
(2 sinx + 1)(sinx - 1) = 0 factor.

- Second, set each factor equal to zero to obtain solutions within the interval [0, 2π):

x = π/2, 7π/6, and 11π/6

*NOTE: when working with a quadratic equation, make sure that the equation involves a single trigonometric function (rewrite if necessary).
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- Squaring and converting to quadratic type:
- If it is necessary to square the equation to convert it into a quadratic type, then remember to check for extraneous solutions!