Can You Solve These Grade 11 Trigonometry Problems?

[Raza]

$$1)~How~would~I~put~this~into~exact~ratio?$$

$$Cos\frac{5pi}{6}$$



$$2)~How~would~factor~this~expression?$$

$$Cosx-2SinxCosx$$



$$This~is~not~a~quick~question~but~I~think~it~is~tough.~I~got~1~out~of~9~in~it.$$
$$3)~How~would~I~solve~this~equation?$$

$$6Cosx^2x+Sinx-4=0~for~ 0^0<x<360^0$$



$$4)~How~many~solutions~does~the~equation~\sqrt 2cos(2x)=1~for~0<x<2pi~have?$$

$$Any~help~will~be~appreciated!$$

$$THANK~YOU$$

 

[whozum]

You give it a try first.

 

[Raza]

I know the answer to the 1st question is -0.86602540378443864676372317075294 but how do I convert this into a fraction? My guess the answer has to be $$Cos(\frac{5pi}{1})$$

As for the 2nd question, my teacher gave me a zero on this question and I don't know what I did wrong.
$$(Cosx-Sinx)(Cosx+Sinx)$$

For the 3rd question,
$$6Cosx^2x+Sinx-4=0~for~ 0^0~<0<360^0$$ factors into $$(3Cosx-2)(3Cosx+2)$$
| |
| |
| |
| |-------------->$$2Cosx+2=0$$
| |-------------->$$Cosx=-\frac{2}{2}$$
| |-------------->$$Cosx=-1$$
|
|
$$3Cosx-2=0$$
$$Cosx=\frac{2}{3}$$

 

[whozum]

-0.86602540378443864676372317075294 reduces to a very common ratio in trig. Do you have a unit circle handy? Or some trig tables? Maybe drawing out a 30-60-90 triangle with hypotenuse 1, short side 0.5. Solve for the long side.

 

[Raza]

Thank you for help me with question 1, it's $$-\frac{\sqrt3}{2}$$

 

[James R]

The factorisation question...

$$\cos x - 2 \sin x \cos x$$

You answered:

$$(\cos x-\sin x)(\cos x+ \sin x)$$

But if we expand that out we get:

$$\cos^2 x - \sin^2 x$$

which is obviously not the same as the original expression. The best you can do for the original expression is:

$$\cos x - 2 \sin x \cos x = \cos x(1 - 2 \sin x)$$

 

[whozum]

If you are allowed double angles, you can get
$$\cos x - \sin 2x$$ which is a bit simpler.

 

[Raza]

I understand how to do 1 and 2 (Thank You) but how would you do the 4th one? That is the one where I got 1 out 9 on the unit test.

 

[whozum]

$$\sqrt 2 \cos 2x = 1$$ for x.

Divide both sides by $\sqrt 2$.

Then for simplicity, say u = 2x.

$$\cos u = \frac{1}{\sqrt{2}} = \frac{\sqrt 2}{2}$$

Solving that gives you u. Then use the u = 2x equation above to find x.

 

[Raza]

I have tried to figure out question 3 myself. Could someone please check my work?

$$6Cos^2x+Sinx-4=0~for~ 0^0$$

$$6Cos^2x+Sinx=4$$

$$6(1-Sin^2x)+Sinx=4$$

$$1-Sin^2x+Sinx=\frac{4}{6}$$ *Rearrange it as Sin²x+1-Sinx and factor it

$$(Sinx+1)(Sinx-1)=\frac{4}{6}$$

For (Sinx-1)
$$Sinx-1=\frac{4}{6}$$
$$Sinx=-\frac{5}{3}$$

For (Sinx+1)
$$Sinx+1=\frac{4}{6}$$
$$Sinx=\frac{5}{3}$$

 

[whozum]

Between lines 3 and 4, you divide both sides by 6, you must also divide the single (sinx) term by 6.

Also, between lines 4 and 5 you factor $1 - sin^2x + sinx$ incorrectly. You seem to have trouble with factoring, try reviewing some. When you expand the binomials you wrote you get [itex]sin^2x - 1 [/tex] which isn't equal to the original expression. <br /> <br /> Also note, sinx is always bounded by -1 and 1. Your two answers lie outside these bounds.[/itex]

 

[Ouabache]

I have tried to figure out question 3 myself. Could someone please check my work?

$6Cos^2x+Sinx-4=0~for~ 0^0$

$6Cos^2x+Sinx=4$

$6(1-Sin^2x)+Sinx=4$
$1-Sin^2x+Sinx=\frac{4}{6}$

I see a problem. On the left side you neglected to divide $\frac{Sinx}{6}$
then you would have: $1-Sin^2x+\frac{Sinx}{6}=\frac{4}{6}$ which isn't too easy to work with.

instead, how about multiplying through and factoring?
$6(1-Sin^2x)+Sinx=4$
$6-6Sin^2x+Sinx-4=0$
$-6Sin^2x+Sinx+2=0$ .. you get the idea?

[edition]: I see whozum spotted the same trouble.. If you factor as I suggest, you will get two solutions that do lie on the unit circle (bounded by +1 and -1)