Alg. II and Trig Questions

[KingAce]

I was out this past week as I had a funeral to attend, and so I missed this week's lesson on trigonometric identities. I was wondering if any of you could help me solve these (these are just a few of my homework questions). I have a conversion sheet to transform the equations, but I'd really appreciate some assistance. Thanks.

Solve over the interval 0<x<360 (inequalities are "Or equal to", as well).

3tan^2x=7secx-5

(1-cosx)/(sinx)=sq.rt(3)

sin2xsinx-cos2xcosx=1

 

[tkhunny]

You should have a cheat sheet with various identities. If you don't have one, get one. I remember my high school textbook. The summary page was page 89. It was a very long time ago. Obviously, it was very useful to be burned into my mind after these many years.

One that should be in it is:

cos(A + B) = cos(A)cos(B)-sin(A)sin(B)

This should lead you to the solution of the third.

 

[galactus]

Solve over the interval 0<x<360 (inequalities are "Or equal to", as well).

3tan^2x=7secx-5

There are different approaches for these things, but you could possibly go this route:

\(\displaystyle \L\\3tan^{2}(x)=7sec(x)-5\)

\(\displaystyle \L\\3\left(\frac{sin^{2}x}{cos^{2}x}\right)=\frac{7}{cosx}-5\)

Multiply by $\displaystyle cos^{2}x$ and get:

\(\displaystyle \L\\3sin^{2}x=7cosx-5cos^{2}x\)

\(\displaystyle \L\\3(1-cos^{2}x)=7cosx-5cos^{2}x\)

\(\displaystyle \L\\3-3cos^{2}x=7cosx-5cos^{2}x\)

\(\displaystyle \L\\2cos^{2}x-7cosx+3=0\)

Now, build a quadratic by letting u=cosx

Solve the quadratic and your solutions in the interval.

 

[soroban]

Hello, KingAce!

Solve over the interval $\displaystyle 0^o\,\leq\,x\,\leq\,360^o$

$\displaystyle 1)\;3\cdot\tan^2x\:=\:7\cdot\sec x\,-\,5$

From the identity: $\displaystyle \:\tan^2\theta\:=\:\sec^2\theta\,-\,1$
. . we get: $\displaystyle \:3(\sec^2x\,-\,1) \:=\:7\cdot\sec x\,-\,5$

We have the quadratic: $\displaystyle \:3\cdot\sec^2x\,-\,7\cdot\sec x\,+\,2\:=\:0$

. . which factors: \(\displaystyle \3\cdot\sec x\,-\,1)(\sec x\,-\,2)\:=\:0\)

We have: $\displaystyle \:3\cdot\sec x\,-\,1\:=\:0\;\;\Rightarrow\;\;\sec x\:=\:\frac{1}{3}$ . . . no real roots

And: $\displaystyle \:\sec x\,-\,2\:=\:0\;\;\Rightarrow\;\;\sec x\:=\:2\;\;\Rightarrow\;\;x \:=\:60^o,\:300^o$

\(\displaystyle 2.\;\L\frac{1\,-\,\cos x}{\sin x}\:=\:\sqrt{3}\)

We note that $\displaystyle \sin x\,\neq\,0$ . . . hence: $\displaystyle \,x\,\neq\,0,\:180^o,\:360^o,\,\cdots\;$ [1]


We have: $\displaystyle \:1\,-\,\cos x\:=\:\sqrt{3}\cdot\sin x$

Square both sides: $\displaystyle \:1\,-\,2\cdot\cos x\,+\,\cos^2x\:=\:3\cdot\sin^2x$

. . Then: $\displaystyle \:1\,-\,2\cdot\cos x\,+\,\cos^2x\:=\:3(1\,-\,\cos^2x)$

. . which simplifies to: $\displaystyle \:2\cdot\cos^2x\,-\,\cos x\,-\,1\:=\:0$

. . which factors: \(\displaystyle \\cos x\,-\,1)(2\cdot\cos x\,+\,1)\:=\:0\)


The first gives us: $\displaystyle \:\cos x\,=\,1\quad\Rightarrow\quad x \:=\:0^o,\,360^o,\,\cdots$ . . . which is not allowed.

The second gives us: $\displaystyle \:\cos x\:=\:-\frac{1}{2}\;\;\Rightarrow\;\;x\:=\:120^o,\:240^o$

But $\displaystyle x\,=\,240^o$ is an extraneous root.

The only solution is: $\displaystyle \:x\:=\:120^o$

$\displaystyle 3.\;\sin2x\cdot\sin x\,-\,\cos2x\cdot\cos x\:=\:1$

If you recognize the identity: $\displaystyle \:\cos(A\,+\,B) \:=\:\cos A\cdot\cos B\,-\,\sin A\cdot\sin B$

. . we have: $\displaystyle \:\cos2x\cdot\cos x\,-\,\sin2x\cdot\sin x\:=\:-1$

Then: $\displaystyle \:\cos3x \:=\:-1\;\;\Rightarrow\;\;3x\:=\:180^o,\:540^o,\:900^o$

Therefore: $\displaystyle \:x\:=\:60^o,\:180^o,\:300^o$