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
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
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
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 :)