PLEASEE HELPPMEEE!!!!
cot2x+sec2x=tan2x+csc2x
and
1+2sinxcosx/sinx+cosx=sinx + cosx
let 2x = y, then
coty + secy = tany + cscy
cosy/siny + 1/cosy = siny/cosy + 1/siny
(cos^2 y + siny)/ sinycosy = sin^2 y + cosy)/sinycosy
cos^2 y + siny = sin^2 y + cosy
cos^2 y - sin^2 y = cosy - siny
(cosy + sinY )(cosy-siny) = cosy - siny
cosy + siny = 1
square both sides
cos^2 y + 2sinycosy + cos^2 y = 1
2sinycosy = 0
sin2y = 0
2y = 0, 180° , 360° , ....
y = 0, 90°, 180°, ...
but period of sin2y is 180°
so other answers are
90+180 = 270
so y = 0, 90, 180, 270, 360, ...
but y = 2x
so x = 0, 45, 90, 135, 180, ...
however, tangent is undefined for 90
cotangent is undefined for 0
csc is undefined for 0, 180, 360
sec is undefined for 90, 270 ...
so there is no solution.
Check your typing.
for the second, I am not sure of your typing,
if you meant it the way you typed it, then .....
1+2sinxcosx/sinx+cosx=sinx + cosx
1 + 2cosx + cosx = sinx + cosx
2cosx = -1
cosx = -1/2
x = 120° or x = 240°
If you meant
(1+2sinxcosx)/(sinx+cosx)=sinx + cosx
(sin^2 x + cos^2 x + 2sinxcosx) /(sinx+cosx) = sinx+cosx
(sinx + cosx)^2/(sinx+cosx) = sinx + cosx
sinx + cosx = sinx + cosx
Ls = RS
this is an identity.
To solve the equations, let's break them down step by step.
1. cot(2x) + sec(2x) = tan(2x) + csc(2x)
This is a trigonometric equation involving cotangent (cot), secant (sec), tangent (tan), and cosecant (csc) functions.
To solve it, we need to simplify and manipulate the equation using trigonometric identities.
First, let's rewrite the equation in terms of sin and cos:
cot(2x) + sec(2x) = tan(2x) + csc(2x)
Using the reciprocal relationship of cot(x) = cos(x)/sin(x) and sec(x) = 1/cos(x), we can substitute:
cos(2x)/sin(2x) + 1/cos(2x) = sin(2x)/cos(2x) + 1/sin(2x)
Next, let's combine the fractions by finding a common denominator. In this case, the denominator will be sin(2x)*cos(2x):
(cos(2x)*cos(2x) + sin(2x))/(sin(2x)*cos(2x)) = (sin(2x) + cos(2x))/(sin(2x)*cos(2x))
Expanding the numerator:
cos^2(2x) + sin(2x) = sin(2x) + cos(2x)
Now, let's simplify further:
cos^2(2x) = cos(2x)
Now, we have a quadratic equation. Rearranging for convenience:
cos^2(2x) - cos(2x) = 0
Factoring out cos(2x):
cos(2x)(cos(2x) - 1) = 0
Now, we can set each factor equal to zero and solve for x.
cos(2x) = 0
2x = pi/2 + n*pi (where n belongs to the set of integers)
x = (pi/4 + n*pi)/2
cos(2x) - 1 = 0
2x = pi + n*pi (where n belongs to the set of integers)
x = (pi/2 + n*pi)/2
Therefore, the solutions for the equation cot(2x) + sec(2x) = tan(2x) + csc(2x) are:
x = (pi/4 + n*pi)/2 and x = (pi/2 + n*pi)/2
Now let's move to the second equation:
2. (1 + 2sin(x)cos(x))/(sin(x) + cos(x)) = sin(x) + cos(x)
This equation involves sine (sin) and cosine (cos) functions.
To solve it, we need to manipulate the equation using trigonometric identities.
First, let's simplify the expression on the left side:
(1 + 2sin(x)cos(x))/(sin(x) + cos(x)) = sin(x) + cos(x)
Next, let's simplify and combine the fractions by finding a common denominator, which in this case is (sin(x) + cos(x)):
(1 + 2sin(x)cos(x))/(sin(x) + cos(x)) = ((sin(x) + cos(x))^2)/(sin(x) + cos(x))
Expanding the numerator:
(1 + 2sin(x)cos(x))/(sin(x) + cos(x)) = (sin(x)^2 + 2sin(x)cos(x) + cos(x)^2)/(sin(x) + cos(x))
Since sin(x)^2 + cos(x)^2 = 1, we have:
(1 + 2sin(x)cos(x))/(sin(x) + cos(x)) = (1 + 2sin(x)cos(x))/(sin(x) + cos(x))
Therefore, the equation is satisfied for all values of x. It is an identity, so no specific solutions are needed.
I hope this helps! Let me know if you have any further questions.