Solving Trigonometric Equations

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Addendum - Solve Sine Trigonometric Equation over the interval [0,2pi)

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Key Questions

  • Examples of trig expressions: # f(x) = sin 2x + cos x; #
    #f(x) = sin x + sin 2x + sin 3x#
    Examples of trig equations: # f(x) = sin 2x + cos x = 0#
    # f(x) = sin x + sin 2x + sin 3x = 0#
    Examples of trig inequalities #f(x) = sin 2x + cos x > 0#
    # f(x) = sin x + sin 2x + sin 3x < 0#

    Use trig Transformation Identities to transform these above trig expressions into trig basic expressions, or expressions in simplest form.
    Example: Transform #f(x) = sin 2x + cos x#. Use Identity #(sin 2a = 2 sin a*cos a)# to transform #f(x).#
    #f(x) = 2*sin x*cos x + cos x = cos x*(2sin x + 1)#
    This is f(x) expressed in simplest form.
    Trig equation in simplest form: #f(x) = cos x*(sin 2x + 1) = 0#
    Trig inequality in simplest form: #f(x) = cos x*(2sinx + 1) > 0#

  • Solving concept. To solve a trig equation, transform it into one, or many, basic trig equations. Solving a trig equation, finally, results in solving various basic trig equations.
    There are 4 main basic trig equations:
    sin x = a; cos x = a; tan x = a; cot x = a.
    Exp. Solve sin 2x - 2sin x = 0
    Solution. Transform the equation into 2 basic trig equations:
    2sin x.cos x - 2sin x = 0
    2sin x(cos x - 1) = 0.
    Next, solve the 2 basic equations: sin x = 0, and cos x = 1.
    Transformation process.
    There are 2 main approaches to solve a trig function F(x).
    1. Transform F(x) into a product of many basic trig functions.
    Exp. Solve F(x) = cos x + cos 2x + cos 3x = 0.
    Solution. Use trig identity to transform (cos x + cos 3x):
    F(x) = 2cos 2x.cos x + cos 2x = cos 2x(2cos x + 1 ) = 0.
    Next, solve the 2 basic trig equations.
    2. Transform a trig equation F(x) that has many trig functions as variable, into a equation that has only one variable. The common variables to be chosen are: cos x, sin x, tan x, and tan (x/2)
    Exp Solve #sin ^2 x + sin^4 x = cos^2 x#
    Solution. Call cos x = t, we get
    #(1 - t^2)(1 + 1 - t^2) = t^2#.
    Next, solve this equation for t.
    Note . There are complicated trig equations that require special transformations.

  • As a general description, there are 3 steps. These steps may be very challenging, or even impossible, depending on the equation.

    Step 1: Find the trigonometric values need to be to solve the equation.
    Step 2: Find all 'angles' that give us these values from step 1.
    Step 3: Find the values of the unknown that will result in angles that we got in step 2.

    (Long) Example
    Solve: #2sin(4x- pi/3)=1#

    Step 1: The only trig function in this equation is #sin#.
    Sometimes it is helpful to make things look simpler by replacing, like this:
    Replace #sin(4x- pi/3)# by the single letter #S#. Now we need to find #S# to make #2S=1#. Simple! Make #S=1/2#
    So a solution will need to make #sin(4x- pi/3)=1/2#

    Step 2: The 'angle' in this equation is #(4x- pi/3)#. For the moment, let's call that #theta#. We need #sin theta = 1/2#
    There are infinitely many such #theta#, we need to find them all.

    Every #theta# that makes #sin theta = 1/2# is coterminal with either #pi/6# or with #(5 pi)/6#. (Go through one period of the graph, or once around the unit circle.)
    So #theta# Which, remember is our short way of writing #4x- pi/3# must be of the form: #theta = pi/6+2 pi k# for some integer #k# or of the form #theta = (5 pi)/6 +2 pi k# for some integer #k#.

    Step 3:
    Replacing #theta# in the last bit of step 2, we see that we need one of: #4x- pi/3 = pi/6+2 pi k# for integer #k#
    or #4x- pi/3 = (5 pi)/6+2 pi k# for integer #k#.

    Adding # pi/3# in the form #(2 pi)/6# to both sides of these equations gives us:
    #4x = (3 pi)/6+2 pi k = pi/2+2 pi k# for integer #k# or
    #4x = (7 pi)/6+2 pi k# for integer #k#.

    Dividing by #4# (multiplying by #1/4#) gets us to:

    #x= pi/8+(2pi k)/4# or
    #x=(7 pi)/24+(2 pi k)/4# for integer #k#.

    We can write this in simpler form:
    #x= pi/8+pi/2 k# or
    #x=(7 pi)/24+pi/2 k# for integer #k#.

    Final note The Integer #k# could be a positive or negative whole number or 0. If #k# is negative, we're actually subtracting from the basic solution.

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Trigonometric Identities and Equations