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Solving Trigonometric Equations Using Basic Algebra

Substitute in potential solutions or solve using inverse trig functions.

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Solving Trigonometric Equations Using Algebra

As Agent Trigonometry, you are given this clue: 2sinx2=0. If 0x<2π, what is/are the value(s) of x.

Guidance

In the previous concept, we verified trigonometric identities, which are true for every real value of x. In this concept, we will solve trigonometric equations. An equation is only true for some values of x.

Example A

Verify that cscx2=0 when x=5π6.

Solution: Substitute in x=5π6 to see if the equations holds true.

csc(5π6)21sin(5π6)2112222=0=0=0=0

This is a true statement, so x=5π6 is a solution to the equation.

Example B

Solve 2cosx+1=0.

Solution: To solve this equation, we need to isolate cosx and then use inverse to find the values of x when the equation is valid. You already did this to find the zeros in the graphing concepts earlier in this chapter.

2cosx+12cosxcosx=0=1=12

So, when is the cosx=12? Between 0x<2π,x=2π3 and 4π3. But, the trig functions are periodic, so there are more solutions than just these two. You can write the general solutions as x=2π3±2πn and x=4π3±2πn, where n is any integer. You can check your answer graphically by graphing y=cosx and y=12 on the same set of axes. Where the two lines intersect are the solutions.

Example C

Solve 5tan(x+2)1=0, where 0x<2π.

Solution: In this example, we have an interval where we want to find x. Therefore, at the end of the problem, we will need to add or subtract π, the period of tangent, to find the correct solutions within our interval.

5tan(x+2)15tan(x+2)tan(x+2)=0=1=15

Using the tan1 button on your calculator, we get that tan1(15)=0.1974. Therefore, we have:

x+2x=0.1974=1.8026

This answer is not within our interval. To find the solutions in the interval, add π a couple of times until we have found all of the solutions in [0,2π].

x=1.8026+π=1.3390=1.3390+π=4.4806

The two solutions are x=1.3390 and 4.4806.

Concept Problem Revisit To solve this equation, we need to isolate sinx and then use inverse to find the values of x when the equation is valid.

2sinx2=02sinxsinx=2=22

So now we need to find the values of x for which sinx=22. We know from the special triangles that this value of sine holds true for a 45 angle, but is that the only value of x for which it is true?

We are told that 0x<2π. Recall that the sine is positive in both the first and second quadrants, so sinx=22 when x also is 135.

Guided Practice

1. Determine if x=π3 is a solution for 2sinx=3.

Solve the following trig equations in the interval 0x<2π.

2. 3cos2x5=0

3. 3sec(x1)+2=0

Answers

1. 2sinπ3=3232=3 Yes, x=π3 is a solution.

2. Isolate the cos2x and then take the square root of both sides. Don’t forget about the ±!

9cos2x59cos2xcos2xcosx=0=5=59=±53

The cosx=53 at x=0.243 rad (use your graphing calculator). To find the other value where cosine is positive, subtract 0.243 from 2π, x=2π0.243=6.037 rad.

The cosx=53 at x=2.412 rad, which is in the 2nd quadrant. To find the other value where cosine is negative (the 3rd quadrant), use the reference angle, 0.243, and add it to π. x=π+0.243=3.383 rad.

3. Here, we will find the solution within the given range, 0x<2π.

3sec(x1)+23sec(x1)sec(x1)cos(x1)=0=2=23=32

At this point, we can stop. The range of the cosine function is from 1 to -1. 32 is outside of this range, so there is no solution to this equation.

Explore More

Determine if the following values for x. are solutions to the equation 5+6cscx=17.

  1. x=7π6
  2. x=11π6
  3. x=5π6

Solve the following trigonometric equations. If no solutions exist, write no solution.

  1. 1cosx=0
  2. 3tanx3=0
  3. 4cosx=2cosx+1
  4. 5sinx2=2sinx+4
  5. secx4=secx
  6. tan2(x2)=3

Sole the following trigonometric equations within the interval 0x<2π. If no solutions exist, write no solution.

  1. cosx=sinx
  2. 3cscx=2
  3. 6sin(x2)=14
  4. 7cosx4=1
  5. 5+4cot2x=17
  6. 2sin2x7=6

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