Question

Please help!! 8. Solve these equations:
a. sin(x + (π / 4)) – sin(x – (π / 4) = 1

b. sin x cos x = √3 / 4
c.tan^2 x–3tanx+2=0

Answer 1

b) i remember that \[\sin x \cos x = \frac 12 \sin (2x)\] does that help at all with this one?

Answer 2

hmm I;m not sure. This is all very hard for me

Answer 3

\[\sin(x) \cos(x)= \frac 12 \sin(2x) = \frac {\sqrt 3}4\]
\[\large \frac 12 \sin(2x) = \frac {\sqrt 3}4\]
\[\large \sin(2x) = \frac {\sqrt 3}4 \times 2\]
\[\large \sin(2x) = \frac {\sqrt 3}2\]
\[\large (2x) = \sin^{-1} (\frac {\sqrt 3}2)\]
\[\large 2x = \sin^{-1} (\frac {\sqrt 3}2)\]
use calculator to work out from here, and tell me ur answer and I'll compare to mine? @washcaps ?

Answer 4

pi/6?

Answer 5

perfect, that's what i got too... x = pi/6

Answer 6

oh so that is the solution to b) then! what about the others?

Answer 7

i just tried a couple of different things for a)... none of them worked... @hartnn can u help explain this please, i'm lost

Answer 8

@ganeshie8 ur awesome at trig, do u have a sec?

Answer 9

I can solve b. but it's kinda long.

Answer 10

@Hero we solved be already

Answer 11

nah, i think we got b, a and c are still a mystery though

Answer 12

@iambatman ... u have a sec? help plz?

Answer 13

c isn't hard. let tanx = y

Answer 14

y^2 - 3y + 2 = 0

Answer 15

ahh, gotcha, then solve using quadratic equation? @Hero ?
that's awesome!

Answer 16

And then after you find the factors set equal to zero, then replace y with tan x again

Answer 17

And finish solving

Answer 18

set it equal to zero

Answer 19

y=2,1

Answer 20

@Hero

Answer 21

Washcaps, what you do is this:

(y - 2)(y - 1) = 0
y - 2 = 0
y - 1 = 0

tan(x) - 2 = 0
tan(x) - 1 = 0

tan(x) = 2
tan(x) = 1

then take the inverse tan of both sides to isolate x

Answer 22

arctan x=2? = -2.18 +- 2pin

Answer 23

For A use this formula:

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

Answer 24

wait I still don't know c

Answer 25

You don't know what x equals

Answer 26

You have to take inverse tan of both sides to get:

\(x = \tan^{-1}(2)\)
\(x = \tan^{-1}(1)\)

Answer 27

x=1.1, pi/4?

Answer 28

Correct

Answer 29

oh really? that's the solution? I thought it was more difficult

Answer 30

Yes, that's the solution for c

Answer 31

Do you know how to solve A)?

Answer 32

For part a apply these formulas:

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

Answer 33

I have an alternative solution for b that uses quadratic formula as well.

Answer 34

I think I'm okay on b because I have an answer already (pi/6). could you walk me through a)? it is my last set of questions like this

Answer 35

There's more than one solution for b though

Answer 36

really?

Answer 37

Yes, really.

Answer 38

ok so I know pi/6 is one of them. what would be the other?

Answer 39

I will use my steps and I will verify what the actual solutions are.

Answer 40

Ok

Answer 41

sin (x) cos (x) = √(3)/ 4

Multiply both sides by 4:
4sin(x)cos(x) = √3

Square both sides: (Remember \((ab)^c = a^cb^c\):
(4sin(x)cos(x))^2 = √(3)^2
16sin^2(x)cos^2(x) = 3
16sin^2(x)(1 - sin^2(x)) = 3

Answer 42

That's what we have so far. Now upon distribution:
16sin^2(x) - 16sin^4(x) = 3

Answer 43

We will do a substitution and let sin^2(x) = y

16y - 16y^2 = 3

Answer 44

so y=1/4, 3/4

Answer 45

@washcaps...no

Answer 46

After rearranging the terms we get:
-16y^2 + 16y - 3 = 0
16y^2 - 16y + 3 = 0

Factoring this we get:
16y^2 -(12 + 4)y + 3 = 0
16y^2 - 12y - 4y + 3 = 0
3y(4y - 3) - 1(4y - 3) = 0
(4y - 3)3y - 1) = 0

4y - 3 = 0
3y - 1 = 0

*Important* ---> We now have to substitute the sin^2x back in for y:

4sin^2x - 3 = 0
3sin^2x - 1 = 0

Answer 47

sin^2x = 3/4
sin^2x = 1/3

(sin(x))^2 = 3/4
(sin(x))^2 = 1/3

\(\sin(x) = \sqrt{\frac{3}{4}}\)

\(\sin(x) = \sqrt{\frac{1}{3}}\)

Answer 48

\(x = \sin^{-1}\left(\sqrt{\frac{3}{4}}\right)\)
\(x = \sin^{-1}\left(\sqrt{\frac{1}{3}}\right)\)

Answer 49

x=pi/3, 0.61

Answer 50

We have to check negative results as well.

Answer 51

how do we do that?

Answer 52

For a) use the identities:
sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
sin(A-B) = sin(A)cos(B) - cos(A)sin(B)
Therefore, sin(A+B) - sin(A-B) = 2cos(A)sin(B)

Answer 53

@ranga, I already posted that earlier

Answer 54

sorry, didn't read the whole thing.

Answer 55

@ranga I'm just not sure how to use the identity though

Answer 56

x = A
pi/4 = B

Answer 57

@Hero can we finish b before we move to a please. I'm getting confused lol

Answer 58

You take the inverse sine of the negative values

\(\sin^{-1}\left(-\sqrt{\frac{3}{4}}\right)\)
\(\sin^{-1}\left(-\sqrt{\frac{1}{3}}\right)\)

Answer 59

The final thing is, now you have to plug the values you found for x back in to the original equation for b to make sure each value works.

Answer 60

x=pi/6, 5pi/3, pi/3, and +- 0.61?

Answer 61

Plug each of those values into b to make sure they work. You'll want to use exact values for this.

Answer 62

You should only have four solutions, not five.

Answer 63

5pi/3 doesn't work. it gives -sqrt 3/4

Answer 64

Where'd you even get that from?

Answer 65

what?

Answer 66

5pi/3

Answer 67

hmm I'm not sure lol. so the answers are pi/6, pi/3 and =- 0.61?

Answer 68

0.6154 to be specific

Answer 69

\(x = \sin^{-1}\left(\sqrt{\frac{3}{4}}\right)\)
\(x = x = \sin^{-1}\left(\sqrt{\frac{1}{3}}\right)\)
\(x = \sin^{-1}\left(-\sqrt{\frac{3}{4}}\right)\)
\(x = x = \sin^{-1}\left(-\sqrt{\frac{1}{3}}\right)\)

Answer 70

That's only four possible results

Answer 71

sin^-1 (-sqrt 3/4) = 5pi/3

Answer 72

Are you entering that into your calculator correctly

Answer 73

yes. also I don't see where the pi/6 answer is

Answer 74

I don't agree that you are entering that into your calculator correctly. Type how you are inputting it

Answer 75

-sqrt(3/4) not -sqrt(3)/4

Answer 76

that's what I typed. Could you just tell me the 4 solutions? I have to leave in 5 minutes and really need this done. You showed me the steps which i understand but arguing over the answers is taking too much time

Answer 77

What calculator are you using?

Answer 78

factoring calculator online

Answer 79

give me the link to the site.

Answer 80

http://www.freemathhelp.com/factoring-calculator.php please hurryy i have to leave

Answer 81

How can you possibly calculate anything using that site? That's not a real calculator

Answer 82

it seems to always work

Answer 83

Explain how to use it

Answer 84

It would be easier if you just tell me what you are inputting into it.

Answer 85

I really don't have time. I just type in the equation and it gives me an answer. symbolab.com said the pi/6 is the only answer too. can you please just give me the 4 solutions. I will be back on in a couple hours

Answer 86

Hang on

Answer 87

http://www.wolframalpha.com/input/?i=solve+sin+x+cos+x+%3D+%E2%88%9A3+%2F+4+

Answer 88

As you can see, there are at least four values for x.

Answer 89

I see that, but can't really read them proper;y. what are they?

Answer 90

+/- 0.61
-pi/3
+pi/3

But the problem is, you have to check them because, from what I recall, only the two positive values work. But double check.

Answer 91

jack1 and I arrived at pi/6 as an answer. read up top

Answer 92

hang on. Let me show you something

Answer 93

http://www.wolframalpha.com/input/?i=sin+%28pi%2F3%29+cos+%28pi%2F3%29+%3D+%E2%88%9A3+%2F+4+

Answer 94

pi/6 is also true

Answer 95

I know

Answer 96

Those are the two correct answers then

Answer 97

pi/3 and pi/6 are the solutions?