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# Solve these equations:

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Solve these equations:

a. sin(x + (π / 4)) – sin(x – (π / 4) = 1

b. sin x cos x = √3 / 4

c. tan2 x – 3tan x + 2 = 0

Apr 7, 2015

#1
+94609
+15

a. sin(x + (π / 4)) – sin(x – (π / 4) = 1

Notice that we can write this in terms of  sum and difference  identities

[sin(x)cos(pi/4) + sin(pi/4)cosx] - [sin(x)cos(pi/4) - sin(pi/4)cosx] = 1

2sin(pi/4) cos(x)  = 1    divide both sides by 2

sin(pi/4)cos(x)  = 1/2      and sin(pi/4)  =  1/√2

So we have

[1/√2]cos(x)  = 1/2     multiply both sides by √2

cos(x)  = √2 / 2   = 1 /√2

And this happens when x = pi/4 and when x = 7pi/4   on the interval [0, 2pi ]

b.  sin(x)cos(x) = √3 / 4

Notice that sin(x)cos(x)  = (1/2)sin(2x)....so we have.

(1/2) sin (2x)  = √3/4      multiply both sides by 2

sin(2x)  = √3/2

sin(x) = √3/2        at pi/3 and at 2pi/3

So

sin(2x)  = √3/2    when x =  pi/6,  pi/3   on [0, 2pi]

Notice, BJ, that  two other solutions on [0,2pi] "work" in the original equation.....7pi/6 and 4pi/3  because sine and cosine are both negative at those values.....but.....multiplying them togeter results in a positive.....

Here's the graph of this one.....https://www.desmos.com/calculator/chm8mhm9ib

c.  tan2 x – 3tan x + 2 = 0     factor

(tan x - 2) (tan x - 1) = 0     set each factor to 0

So....

tan x  - 2   = 0    add 2 to both sides

tan x = 2    and this happens at about 63.43°  and at about [180 + 63.43]°  = 243.43°  on  [0, 360]degrees

And the other solution is

tan x  - 1 = 0   add 1 to both sides

tan x = 1    and this happens at 45°  and 225°   on [0, 360 ] degrees

Apr 7, 2015

#1
+94609
+15

a. sin(x + (π / 4)) – sin(x – (π / 4) = 1

Notice that we can write this in terms of  sum and difference  identities

[sin(x)cos(pi/4) + sin(pi/4)cosx] - [sin(x)cos(pi/4) - sin(pi/4)cosx] = 1

2sin(pi/4) cos(x)  = 1    divide both sides by 2

sin(pi/4)cos(x)  = 1/2      and sin(pi/4)  =  1/√2

So we have

[1/√2]cos(x)  = 1/2     multiply both sides by √2

cos(x)  = √2 / 2   = 1 /√2

And this happens when x = pi/4 and when x = 7pi/4   on the interval [0, 2pi ]

b.  sin(x)cos(x) = √3 / 4

Notice that sin(x)cos(x)  = (1/2)sin(2x)....so we have.

(1/2) sin (2x)  = √3/4      multiply both sides by 2

sin(2x)  = √3/2

sin(x) = √3/2        at pi/3 and at 2pi/3

So

sin(2x)  = √3/2    when x =  pi/6,  pi/3   on [0, 2pi]

Notice, BJ, that  two other solutions on [0,2pi] "work" in the original equation.....7pi/6 and 4pi/3  because sine and cosine are both negative at those values.....but.....multiplying them togeter results in a positive.....

Here's the graph of this one.....https://www.desmos.com/calculator/chm8mhm9ib

c.  tan2 x – 3tan x + 2 = 0     factor

(tan x - 2) (tan x - 1) = 0     set each factor to 0

So....

tan x  - 2   = 0    add 2 to both sides

tan x = 2    and this happens at about 63.43°  and at about [180 + 63.43]°  = 243.43°  on  [0, 360]degrees

And the other solution is

tan x  - 1 = 0   add 1 to both sides

tan x = 1    and this happens at 45°  and 225°   on [0, 360 ] degrees

CPhill Apr 7, 2015
#2
+95361
+5

That looks like a lot of work Chris :)

Do you understand all that Brittany - I guess if you don't you will soon say so :)

Apr 8, 2015
#3
+3691
+5

Absolutely, melody! I understand it. and yes, I would let you guys know if I didn't understand it. Thanks for checking tho... :)

Apr 11, 2015
#4
+95361
+5

Thanks  BrittanyJ,

Did you give CPhill points - looks like you might have forgotten

Apr 11, 2015