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M1410503Part1 - (Section 5.3 Solving Trig Equations 5.22...

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(Section 5.3: Solving Trig Equations) 5.22 SECTION 5.3: SOLVING TRIG EQUATIONS PART A: BASIC EQUATIONS IN sin, cos, csc, OR sec (LINEAR FORMS) Example Solve: 5cos x 2 = 3cos x (It is assumed that you are to give all real solutions and to give them in exact form – no approximations – unless otherwise specified.) Conditional Equations This is an example of a conditional equation. It is true (i.e., the left side equals the right side) for some real values of x but not for others. In other words, the truth of the equation is conditional, depending on the particular real value that x takes on. You should be used to solving conditional equations in your Algebra courses. This is different from an identity, which holds true for all real values of x (for instance) for which all expressions involved are defined as real quantities. An identity may be thought of as an equation that has as its solution set the intersection (overlap) of the domains of the expressions involved. Solution First , solve for cos x . This process is no different from solving the linear equation 5 u 2 = 3 u for u . In fact, you could employ the substitution u = cos x and do exactly that. 5cos x 2 = 3cos x 2cos x = 2 cos x = 2 2
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(Section 5.3: Solving Trig Equations) 5.23 We want to find all angles whose cos value is 2 2 . We will use radian measure, which corresponds to “real number” solutions for x . Second , because cos x has period 2 π , we will first find solutions in the interval 0, 2 π ) . Later, we will find all of their coterminal “twin” angles. If you are more comfortable with “slightly negative” Quadrant IV angles such as π 4 than angles such as 7 π 4 , then you may want to look in the interval π 2 , 3 π 2 , instead. Is there an “easy” angle x whose cos value is 2 2 ? Yes, namely π 4 , which is cos 1 2 2 . Look at the Unit Circle. Look at the point corresponding to the π 4 angle. It turns out that there is another point on the Unit Circle that has the same horizontal (or what we used to call “ x ”) coordinate, 2 2 , so we must look for another angle with that same cos value of 2 2 . We know that this point lies
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