Question

solve equation in interval from 0 to 2 pi
7 cos 2t=3

Answer 1

\[\huge \cos2t=\cos^2 t-\sin^2 t\]

Answer 2

and I still don't know what to do next

Answer 3

I rewrote cos2t which means that you should use your logic and use the reqritten expression of cos2t in your equation instead.

Answer 4

And then \[\huge \cos^2 t -\sin^2 t=2\cos^2 t-1\]

Answer 5

rewritten*

Answer 6

I am totally lost

Answer 7

\[7\cos2t=3\]
That is your given equation.
Now I rewrite it using what I put up above.
\[\large 7(\cos^2 t -\sin^2 t)=3\]
Then I rewrite the expression in the brackets again. I'm literally doing this step by step for your own sake and benefit.
\[\huge 7(\cos^2 t -(1-\cos^2 t))=3\]
I rewrote sin^2 t using my trig identities.
\[\cos^2 t + \sin^2 t=1\]
Rearrange that trig identity I find myself with this:
\[\sin^2 t=1-\cos^2 t\]

Answer 8

Then
\[\huge 7(\cos^2 t-1+\cos^2 t)=3\]

Answer 9

\[\huge 7(2\cos^2 t -1)=3\]

Answer 10

Now expand/distribute the LHS.

Answer 11

14cos^2t-7=3

Answer 12

14cos^2t=10

Answer 13

is that right so far

Answer 14

I don't see where this is getting me. My new equation looks very much like the original one

Answer 15

That's correct. That's not the original equation.
\[\huge 14\cos^2t=10\]

Answer 16

\[\huge \cos 2t\neq \cos^2 t\]

Answer 17

Move the 14 to the RHS. by dividing both sides by 14.

Answer 18

\[\huge \cos^2 t=\frac{10}{14}\]

Answer 19

Simplify the fraction on the RHS. And then find cost. There will be two solutions since something that's to the power of 2 must have two solutions. One must be negative and the other must be positive.

Answer 20

5/7

Answer 21

Yep now find cost.

Answer 22

by squaring both sides.

Answer 23

cos of 5/7?

Answer 24

No. I told you to square root both sides.

Answer 25

\[\huge \sqrt{\cos^2 t}=\sqrt{\frac{5}{7}}\]

Answer 26

25/49?

Answer 27

square root does not equal squaring.

Answer 28

right

Answer 29

So you understand that if you square root 1 what would your answer be. This is an example for you to understand that every time you're square rooting something, there should be two solutions.
\[\sqrt{1}=?\]

Answer 30

what would the answer(s) be when you square root 1?

Answer 31

= and -

Answer 32

+ and -

Answer 33

So what would cost equal in this case?
\[\cos t=?\]

Answer 34

\[\huge \cos t=\pm\sqrt{\frac{5}{7}}\]

Answer 35

Okay now, we find t.

Answer 36

If there's a +/- sing in front of it, then using logic, there must be 4 solutions for t. One in each quadrant.

Answer 37

sign*

Answer 38

So what's t equal to? Get your calculator out.

Answer 39

I have it

Answer 40

So did you get
t=32*19', 147*41', 212*19' and 327*41'?

Answer 41

the asterisk is meant to be degrees signs.

Answer 42

no what did you enter into cal to get those numbers

Answer 43

I entered:
\[\huge \cos^{-1}\sqrt{\frac{5}{7}}=32.3115...\]

Answer 44

Then I converted that number into degrees and minutes.

Answer 45

Then I started getting values for the second, third and fourth quadrant.

Answer 46

so 180-32.311 and 180+32.31 and 360-32.31...

Answer 47

all those values I got were converted to degrees and minutes

Answer 48

But if you didn't convert it and just got the values and put the degrees sign up, then you're still correct. But converting it into degrees and minutes is a better answer in maths.

Answer 49

ok this is really a lot to take in. I need to study this. thanks for doing it step by step. very confusing. do not like this section of algebra

Answer 50

This is trigonometry for you mate. Work on it. You will get better with time and practice like you said. No worries and good luck.