1)sinx+cosx=1
2)sinx+cosx=1+sin2x
Are we solving ? I will assume domain 0 ≤ x≤ 2π
first one:
square both sides
sin^2 x + 2sinxcosx + cos^2 x = 1
1 + sin 2x = 1
sin2x = 0
2x = 0 , π, 2π , 3π, 4π
x = 0 , π/2, π , 3π/2 2π
but since we squared , we have to verify all answers
for x=0
LS = sin) + cos) = 0+1 = 1 = RS
for x=π/2
sinπ/2 + cosπ/2 = 1/√2 + 1/√2 = 2/√2 ≠ RS
for x= π
sinπ + cosπ = 0-1 ≠ RS
for x= 3π/2
LS = .......≠ RS
for x=2π
LS = sin2π + cos2π = 0 + 1 = 1 = RS
so x = 0 or x = 2π
also square both sides
sin^2 x + 2sinxcosx + cos^2 x = 1 + 2sin2x + sin^2 (2x)
2sinxcosx = 2sin2x + sin^2 (2x)
sin 2x = 2 sin 2x + sin^2(2x)
sin^2(2x) + sin2x = 0
sin2x(sin2x + 1) = 0
sin2x = 0 or sin2x = -1
2x = 0,π,2π,3π,4π or 2x = 3π/2, 7π/2
x = 0,π/2,π,3π/2,2π or x = 3π/2,7π/4
The above are potential solutions, I will leave it up to you to verify which are actual solutions.
Remember, we squared , so all solutions must be verified.
1) To solve the equation sin(x) + cos(x) = 1, we can use some trigonometric identities.
The first step is to rewrite the equation in terms of a single trigonometric function. We can use the identity sin^2(x) + cos^2(x) = 1, which states that the sum of the squares of sin(x) and cos(x) is always equal to 1. Rearranging the equation, we have sin^2(x) = 1 - cos^2(x).
Substituting this into the original equation, we get:
1 - cos^2(x) + cos(x) = 1
Rearranging further, we have:
cos^2(x) - cos(x) = 0
Now, we can factor out the common factor of cos(x):
cos(x)(cos(x) - 1) = 0
To find the solutions, we set each factor equal to zero:
cos(x) = 0 or cos(x) - 1 = 0
Solving for x in each case, we have:
For cos(x) = 0, x = π/2 or 3π/2 (where cos(x) equals zero).
For cos(x) - 1 = 0, x = 0 (where cos(x) equals 1).
Therefore, the solutions to the equation sin(x) + cos(x) = 1 are x = π/2, 0, and 3π/2.
2) To solve the equation sin(x) + cos(x) = 1 + sin(2x), we can proceed as follows:
Starting with the left side of the equation, we can rewrite it using the double-angle identity for sine, which states that sin(2x) = 2sin(x)cos(x). Substituting this in, we get:
sin(x) + cos(x) = 1 + 2sin(x)cos(x)
Next, rearrange the equation to obtain a quadratic equation in terms of sin(x) and cos(x):
2sin(x)cos(x) - sin(x) - cos(x) + 1 = 0
Factoring by grouping, we have:
(sin(x) - 1)(2cos(x) + 1) = 0
Setting each factor equal to zero, we get:
sin(x) - 1 = 0 or 2cos(x) + 1 = 0
For sin(x) - 1 = 0, solving for x gives x = π/2 (where sin(x) equals 1).
For 2cos(x) + 1 = 0, solving for x gives cos(x) = -1/2. This occurs at x = 2π/3 and 4π/3 (where cos(x) equals -1/2).
Thus, the solutions to the equation sin(x) + cos(x) = 1 + sin(2x) are x = π/2, 2π/3, and 4π/3.