Solve these equations:

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

b. sin *x* cos *x* = √3 / 4

c. tan^{2} *x* – 3tan *x* + 2 = 0

BrittanyJ
Apr 7, 2015

#1**+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. tan^{2} *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

#1**+15 **

Best Answer

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. tan^{2} *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**+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 :)

Melody
Apr 8, 2015