solving trig equations; finding bearing of plane; etc.

stairwy2heavn

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Apr 5, 2007
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1
solve each for x between 0 and 360

5) tan 2θ + tan θ = 0
do i replace tan 2θ with 2tanθ/1-tan^2(x)??
so far I did and I can't seem to solve the problem



7) 2sinθ/3 = -1
sinθ/3 = -1/2
θ/3 = 210 and 330
how do i solve for just θ?


I have no idea how to start for these next two problems!! please help
8) sin 3x - sinx = 2cos 2x
9) sin 2x = 5cos^2(x)
 
Are you unable even to get started on any of these? If so, you probably need more help than we can provide. In hopes that this is not the case....

1) What are you supposed to do with this? (Providing the complete exercise, which includes the instructions, is generally helpful. That's why the "Read Before Posting" thread requested this.

2) Same question as (1).

3) I will guess that you're supposed to solve this, perhaps over some interval. I will also guess that "(cos^2)x" actually means "cos^2(x)" (that is, the square of the cosine of x).

I would suggest applying a double-angle identity, and then applying the Quadratic Formula. Then solve the resulting (linear) trig equations.

4) I would divide through by the cosine, divide through by the -sqrt[3], and then apply the basic reference angle information. (You've memorized that table, right?)

5) Apply the double-angle formula, and then see if you can rearrange this to be able to apply the Quadratic Formula. Then solve the resulting (linear) trig equations.

6) Divide through by the cosine, and then solve using the tangent information. Note that you will be solving for "2theta = (whatever)", so the final solution will require that you first divide off the 2.

7) Your formatting is ambiguous. Please reply using grouping symbols (and showing your work so far) for clarification.

8) This works similarly to (3) and (5).

9) This might require the use of inverse-trig functions. Have you studied those yet?

10) Multiply through by 2, apply the double-angle formula in reverse, and solve the resulting sine equation. Then divide through to get the final answer.

11) Draw the triangle (not necessarily to scale), and label the three sides. Label the one angle you need to find. Apply the Law of Cosines to find the value of the angle. Then use your book's definition of "bearing" to find the final answer.

Eliz.
 
Hello, stairwy2heavn!

I can help on #8 and #9 . . .


\(\displaystyle 8)\;\sin3x\,-\,\sin x \:= \:2\cdot\cos2x\)

Are you familiar with the sum-to-product identities?

. . One of them is: \(\displaystyle \:\sin(A)\,-\,\sin(B)\:=\:2\cdot\cos\left(\frac{A\,+\,B}{2}\right)\cdot\sin\left(\frac{A\,-\,B}{2}\right)\)

The left side is: \(\displaystyle \:\sin3x\,-\,\sin x\:=\:2\cdot\cos\left(\frac{3x\,+\,x}{2}\right)\cdot\sin\left(\frac{3x\,-\,x}{2}\right)\:=\:2\cdot\cos2x\cdot\sin x\)

The equation becomes: \(\displaystyle \:2\cdot\cos2x\cdot\sin x \:=\:2\cdot\cos2x\)

. . . . . . . . . . \(\displaystyle 2\cdot\cos2x\cdot\sin x\,-\,2\cdot\cos2x\:=\:0\)

Factor: \(\displaystyle \:2\cdot\cos2x\cdot(\sin x\,-\,1)\:=\:0\)


And we have two equations to solve:

. . \(\displaystyle \cos2x \:=\:0\;\;\Rightarrow\;\;2x \:=\:90^o,\,270^o,\,450^o,\,630^o\;\;\Rightarrow\;\;\fbox{x\:=\:45^o,\,135^o,\,225^o,\,315^o}\)

. . \(\displaystyle \sin x\,-\,1\:=\:0\;\;\Rightarrow\;\;\sin x\:=\:1\;\;\Rightarrow\;\;\fbox{x\,=\,90^o}\)



\(\displaystyle 9)\;\sin2x \:= \:5\cdot\cos^2x\)

We'll use the identity: \(\displaystyle \:\sin2x \:=\:2\cdot\sin x\cdot\cos x\)

So we have: \(\displaystyle \:2\cdot\sin x\cdot\cos x \:=\:5\cdot\cos^2x\)

. . .\(\displaystyle 2\cdot\sin x\cdot\cos x\,-\,5\cdot\cos^2x\:=\:0\)

Factor: \(\displaystyle \:\cos x\cdot(2\cdot\sin x\,-\,5\cos x) \:=\:0\)


And we have two equations to solve:

. . \(\displaystyle \cos x\:=\:0\;\;\Rightarrow\;\;\fbox{x \:=\:90^o,\,270^o}\)

. . \(\displaystyle 2\cdot\sin x\,-\,5\cos x\:=\:0\;\;\Rightarrow\;\;2\cdot\sin x\:=\:5\cdot\cos x\;\;\Rightarrow\;\;\frac{\sin x}{\cos x} \:=\:\frac{5}{2}\;\;\Rightarrow\;\;\tan x\:=\:\frac{5}{2}\)
. . Then: \(\displaystyle \:\fbox{x \:\approx\:68.2^o,\,248.2^o}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

In all trig equation, check for extraneous roots.
[As it turns out, all these answers are okay.]
 
stairwy2heavn said:
solve each for x between 0 and 360

5) tan 2θ + tan θ = 0
do i replace tan 2θ with 2tanθ/1-tan^2(x)??
so far I did and I can't seem to solve the problem

I'd do:
sin2θ/cos2θ + sinθ/cosθ = 0
2sinθcosθ/(cos^2θ-sin^2θ) + sinθ/cosθ = 0
(3sinθcos^2θ - sin^3θ)/cosθ(cos^2θ-sin^2θ) = 0
sinθ(3cos^2θ - sin^2θ)/cosθ(cos^2θ-sin^2θ) = 0

sinθ = 0
θ = 0, pi, 2pi

or

3cos^2θ - sin^2θ = 0
3 - 4sin^2θ = 0
sin^2θ = 3/4
sinθ = +- sqrt(3)/2
θ = pi/3, 2/3pi, 4/3pi, 5/3pi
 
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