9.5 Solving trigonometric equations  (Page 3/10)

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Solve trigonometric equations using a calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Using a calculator to solve a trigonometric equation involving sine

Use a calculator to solve the equation $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta =0.8,$ where $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in radians.

Make sure mode is set to radians. To find $\text{\hspace{0.17em}}\theta ,$ use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\text{\hspace{0.17em}}$ function. What is shown on the screen is ${\mathrm{sin}}^{-1}\left(\text{\hspace{0.17em}}.$ The calculator is ready for the input within the parentheses. For this problem, we enter $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(0.8\right),$ and press ENTER. Thus, to four decimals places,

${\mathrm{sin}}^{-1}\left(0.8\right)\approx 0.9273$

The solution is

$\theta \approx 0.9273±2\pi k$

The angle measurement in degrees is

$\begin{array}{ccc}\hfill \theta & \approx & 53.1°\hfill \\ \hfill \theta & \approx & 180°-53.1°\hfill \\ & \approx & 126.9°\hfill \end{array}$

Using a calculator to solve a trigonometric equation involving secant

Use a calculator to solve the equation $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}\theta =-4,$ giving your answer in radians.

We can begin with some algebra.

$\begin{array}{ccc}\hfill \mathrm{sec}\text{\hspace{0.17em}}\theta & =& -4\hfill \\ \hfill \frac{1}{\mathrm{cos}\text{\hspace{0.17em}}\theta }& =& -4\hfill \\ \hfill \mathrm{cos}\text{\hspace{0.17em}}\theta & =& -\frac{1}{4}\hfill \end{array}$

Check that the MODE is in radians. Now use the inverse cosine function.

$\begin{array}{ccc}\hfill {\mathrm{cos}}^{-1}\left(-\frac{1}{4}\right)& \approx & 1.8235\hfill \\ \hfill \theta & \approx & 1.8235+2\pi k\hfill \end{array}$

Since $\text{\hspace{0.17em}}\frac{\pi }{2}\approx 1.57\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\pi \approx 3.14,$ 1.8235 is between these two numbers, thus $\text{\hspace{0.17em}}\theta \approx \text{1}\text{.8235}\text{\hspace{0.17em}}$ is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See [link] .

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is $\text{\hspace{0.17em}}\theta \text{​}\text{​}\text{'}\approx \pi -\text{1}\text{.8235}\approx \text{1}\text{.3181}\text{.}\text{\hspace{0.17em}}$ The other solution in quadrant III is $\text{\hspace{0.17em}}\theta \text{​}\text{​}\text{'}\approx \pi +\text{1}\text{.3181}\approx \text{4}\text{.4597}\text{.}$

The solutions are $\text{\hspace{0.17em}}\theta \approx 1.8235±2\pi k\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \approx 4.4597±2\pi k.$

Solve $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =-0.2.$

$\theta \approx 1.7722±2\pi k\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \approx 4.5110±2\pi k$

Solving trigonometric equations in quadratic form

Solving a quadratic equation    may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}u.\text{\hspace{0.17em}}$ If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Solving a trigonometric equation in quadratic form

Solve the equation exactly: $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta +3\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta -1=0,0\le \theta <2\pi .$

We begin by using substitution and replacing cos $\theta \text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ It is not necessary to use substitution, but it may make the problem easier to solve visually. Let $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =x.\text{\hspace{0.17em}}$ We have

${x}^{2}+3x-1=0$

The equation cannot be factored, so we will use the quadratic formula     $\text{\hspace{0.17em}}x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}.$

$\begin{array}{ccc}\hfill x& =& \frac{-3±\sqrt{{\left(-3\right)}^{2}-4\left(1\right)\left(-1\right)}}{2}\hfill \\ & =& \frac{-3±\sqrt{13}}{2}\hfill \end{array}$

Replace $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta ,$ and solve.

$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\theta & =& \frac{-3±\sqrt{13}}{2}\hfill \\ \hfill \theta & =& {\mathrm{cos}}^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\hfill \end{array}$

Note that only the + sign is used. This is because we get an error when we solve $\text{\hspace{0.17em}}\theta ={\mathrm{cos}}^{-1}\left(\frac{-3-\sqrt{13}}{2}\right)\text{\hspace{0.17em}}$ on a calculator, since the domain of the inverse cosine function is $\text{\hspace{0.17em}}\left[-1,1\right].\text{\hspace{0.17em}}$ However, there is a second solution:

$\begin{array}{ccc}\hfill \theta & =& {\mathrm{cos}}^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\hfill \\ & \approx & 1.26\hfill \end{array}$

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
hii
Amit
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Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
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Rubben
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Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
please I need help in maths
Okey tell me, what's your problem is?
Navin