20.2: Equations involving trigonometric functions
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The previous section showed how to solve the basic trigonometric equations
sin(x)=c,cos(x)=c, and tan(x)=c
The next examples can be reduced to these basic equations.
Solve for x.
- 2sin(x)−1=0
- sec(x)=−√2
- 7cot(x)+3=0
Solution
- Solving for sin(x), we get
2sin(x)−1=0(+1)⟹2sin(x)=1(÷2)⟹sin(x)=12
One solution of sin(x)=12 is sin−1(12)=π6. The general solution is
x=(−1)n⋅π6+nπ, where n=0,±1,±2,…
- Recall that sec(x)=1cos(x). Therefore,
sec(x)=−√2⟹1cos(x)=−√2(reciprocal)⟹cos(x)=−1√2=−√22
A special solution of cos(x)=−√22 is
cos−1(−√22)=π−cos−1(√22)=π−π4=4π−π4=3π4
The general solution is
x=±3π4+2nπ, where n=0,±1,±2,…
- Recall that cot(x)=1tan(x). So
7cot(x)+3=0(−3)⟹7cot(x)=−3(÷7)⟹cot(x)=−37⟹1tan(x)=−37(reciprocal)⟹tan(x)=−73
The solution is
x=tan−1(−73)+nπ≈−1.166+nπ, where n=0,±1,±2,…
For some of the more advanced problems it can be helpful to substitute u for a trigonometric expression first, then solve for u, and finally apply the rules from the previous section to solve for the wanted variable. This method is used in the next example.
Solve for x.
- tan2(x)+2tan(x)+1=0
- 2cos2(x)−1=0
Solution
- Substituting u=tan(x), we have to solve the equation
u2+2u+1=0(factor)⟹(u+1)(u+1)=0⟹u+1=0(−1)⟹u=−1
Resubstituting u=tan(x), we have to solve tan(x)=−1. Using the fact that tan−1(−1)=−tan−1(1)=−π4, we have the general solution
x=−π4+nπ, where n=0,±1,±2,…
- We substitute u=cos(x), then we have
2u2−1=0(+1)⟹2u2=1(÷2)⟹u2=12⟹u=±√12=±1√2=±√22⟹u=+√22 or u=−√22
For each of the two cases we need to solve the corresponding equation after replacing u=cos(x).
cos(x)=√22 with cos−1(√22)=π4⟹x=±π4+2nπ where n=0,±1,±2,…
cos(x)=−√22 with cos−1(−√22)=π−cos−1(√22)=π−π4=3π4⟹x=±3π4+2nπ where n=0,±1,±2,…
Thus, the general solution is
x=±π4+2nπ, or x=±3π4+2nπ, where n=0,±1,±2,…
Solve the equation with the calculator. Approximate the solution to the nearest thousandth.
- 2sin(x)=4cos(x)+3
- 5cos(2x)=tan(x)
Solution
- We rewrite the equation as 2sin(x)−4cos(x)−3=0, and use the calculator to find the graph of the function f(x)=2sin(x)−4cos(x)−3. The zeros of the function f are the solutions of the initial equation. The graph that we obtain is displayed below.
The graph indicates that the function f(x)=2sin(x)−4cos(x)−3 is periodic. This can be confirmed by observing that both sin(x) and cos(x) are periodic with period 2π, and thus also f(x).
f(x+2π)=2sin(x+2π)−4cos(x+2π)−3=2sin(x)−4cos(x)−3=f(x)
The solution of f(x)=0 can be obtained by finding the “zeros,” that is by pressing 2ndtrace2, then choosing a left- and right-bound, and making a guess for the zero. Repeating this procedure gives the following two approximate solutions within one period.
The general solution is thus
x≈1.842+2nπ or x≈3.513+2nπ, where n=0,±1,±2,…
- We rewrite the equation as 5cos(2x)−tan(x)=0 and graph the function f(x)=5cos(2x)−tan(x) in the standard window.
To get a better view of the function we zoom to a more appropriate window.
Note again that the function f is periodic. The period of cos(2x) is 2π2=π (see definition [DEF:amplitude-period-phase] on page ), and the period of tan(x) is also π (see equation [EQU:tan-period] on page ). Thus, f is also periodic with period π. The solutions in one period are approximated by finding the zeros with the calculator.
The general solution is given by any of these numbers, with possibly an additional shift by any multiple of π.
x≈1.788+nπ or x≈2.224+nπ or x≈3.842+nπ where n=0,±1,±2,±3,…