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Solve for θ :

sin θ = 0 , 3
  1. We look at the sign of the trigonometric function. sin θ is given as a positive amount ( 0 , 3 ). Reference to the CAST diagram shows that sine is positive in the first and second quadrants.

    S A
    T C

  2. The small angle θ is the angle returned by the calculator:

    sin θ = 0 , 3 θ = arcsin 0 , 3 θ = 17 , 46
  3. Our solution lies in quadrants I and II. We therefore use θ and 180 - θ , and add the 360 · n for the periodicity of sine.

    180 - θ θ
    180 + θ 360 - θ

    I : θ = 17 , 46 + 360 · n , n Z II : θ = 180 - 17 , 46 + 360 · n , n Z = 162 , 54 + 360 · n , n Z

    This is called the general solution .

  4. We can then find all the values of θ by substituting n = ... , - 1 , 0 , 1 , 2 , ... etc. For example,If n = 0 , θ = 17 , 46 ; 162 , 54 If n = 1 , θ = 377 , 46 ; 522 , 54 If n = - 1 , θ = - 342 , 54 ; - 197 , 46 We can find as many as we like or find specific solutions in a given interval by choosing more values for n .

General solution using periodicity

Up until now we have only solved trigonometric equations where the argument (the bit after the function, e.g. the θ in cos θ or the ( 2 x - 7 ) in tan ( 2 x - 7 ) ), has been θ . If there is anything more complicated than this we need to be a little more careful. Let us try to solve tan ( 2 x - 10 ) = 2 , 5 in the range - 360 x 360 . We want solutions for positive tangent so using our CAST diagram we know to look in the 1 st and 3 rd quadrants. Our calculator tells us that arctan ( 2 , 5 ) = 68 , 2 . This is our reference angle. So to find the general solution we proceed as follows:

tan ( 2 x - 10 ) = 2 , 5 [ 68 , 2 ] I : 2 x - 10 = 68 , 2 + 180 · n 2 x = 78 , 2 + 180 · n x = 39 , 1 + 90 · n , n Z

This is the general solution. Notice that we added the 10 and divided by 2 only at the end. Notice that we added 180 · n because the tangent has a period of 180 . This is also divided by 2 in the last step to keep the equation balanced. We chose quadrants I and III because tan was positive and we used the formulae θ in quadrant I and ( 180 + θ ) in quadrant III. To find solutions where - 360 < x < 360 we substitue integers for n :

  • n = 0 ; x = 39 , 1 ; 129 , 1
  • n = 1 ; x = 129 , 1 ; 219 , 1
  • n = 2 ; x = 219 , 1 ; 309 , 1
  • n = 3 ; x = 309 , 1 ; 399 , 1 (too big!)
  • n = - 1 ; x = - 50 , 9 ; 39 , 1
  • n = - 2 ; x = - 140 , 1 ; - 50 , 9
  • n = - 3 ; x = - 230 , 9 ; - 140 , 9
  • n = - 4 ; x = - 320 , 9 ; - 230 , 9

Solution: x = - 320 , 9 ; - 230 ; - 140 , 9 ; - 50 , 9 ; 39 , 1 ; 129 , 1 ; 219 , 1 and 309 , 1

Linear trigonometric equations

Just like with regular equations without trigonometric functions, solving trigonometric equations can become a lot more complicated. You should solve these just like normal equations to isolate a single trigonometric ratio. Then you follow the strategy outlined in the previous section.

Write down the general solution for 3 cos ( θ - 15 ) - 1 = - 2 , 583

  1. 3 cos ( θ - 15 ) - 1 = - 2 , 583 3 cos ( θ - 15 ) = - 1 , 583 cos ( θ - 15 ) = - 0 , 5276 . . . reference angle : ( θ - 15 ) = 58 , 2 II : θ - 15 = 180 - 58 , 2 + 360 · n , n Z θ = 136 , 8 + 360 · n , n Z III : θ - 15 = 180 + 58 , 2 + 360 · n , n Z θ = 253 , 2 + 360 · n , n Z

Quadratic and higher order trigonometric equations

The simplest quadratic trigonometric equation is of the form

sin 2 x - 2 = - 1 , 5

This type of equation can be easily solved by rearranging to get a more familiar linear equation

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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