How can I solve the equation cos x + 1 = sinx?

Answer 1

cos x + 1 = sin x

x= π/2 + (π/2)k where k is a member of the set of integers

First subtract cos x from both sides of the equation.

1 = sin x - cos x

Next, square both sides of the equation.

1 = sin² x - 2sinxcosx + cos² x

Move around parts of the equation to get:

1= sin² x + cos² x - sin(2x)

Since sin² x + cos² x is always equal to 1,

1 = 1 - sin(2x)

Subtract 1 from both sides.

0= -sin(2x)

sin(2x) = 0

2x = 0, π, 2π etc.

x = 0, π/2, π, 3π/2, 2π

Or to include everything:

x = π/2 + (π/2)k where k is a member of the set of integers.

Answer 2

I'm going to assume that 0 <= x < 2pi.

Move sinx to the left hand side, and 1 to the right hand side.

cos(x) - sin(x) = -1

Now, square both sides.

[cos(x) - sin(x)]^2 = (-1)^2

cos^2(x) - 2sin(x)cos(x) + sin^2(x) = 1

Rearrange the terms (this will serve a purpose in the next step)

cos^2(x) + sin^2(x) - 2sin(x)cos(x) = 1

Note the identity, "sine squared x plus cos squared x equals 1"

1 - 2sin(x)cos(x) = 1
-2sin(x)cos(x) = 0
sin(x)cos(x) = 0

Therefore, sin(x) = 0 and cos(x) = 0, and we have to solve for x.

if sin(x) = 0, and 0 <= x < 2pi, then
x = 0, pi

if cos(x) = 0, and 0 <= x < 2pi, then
x = pi/2, 3pi/2

So all possible values of x are
x = 0, pi, pi/2, 3pi/2

However, squaring the original equation may have unnecessarily added values, so we have to check each one.

Test x = 0:

cos(0) + 1 = sin(0) ?
1 + 1 = 0, which is false, so we discard x = 0.

Test x = pi

cos(pi) + 1 = sin(pi) ?
-1 + 1 = 0?
0 = 0, so we keep this solution.

Test x = pi/2

cos(pi/2) + 1 = sin(pi/2) ?
0 + 1 = 1
1 = 1, so we keep this solution.

Test x = 3pi/2

cos(3pi/2) + 1 = sin(3p/2) ?
0 + 1 = -1?
1 = -1 is false, so we discard.

Therefore, x = pi/2, pi

Answer 3

Without using the sum formulas, just write sin(x) as sqrt(1-cos^2(x))

Just square both sides of the equation, simplify, and factor. This becomes

2Cos(x)(Cos(x) + 1) = 0

So, Cos(x) = 0 or Cos(x) = -1, or x = pi/2 or pi

Answer 4

Let y = cos x. Then sin x = ±sqrt(1 - y^2).
So we get y + 1 = ±sqrt(1-y^2).
So (y+1)^2 = 1-y^2
y^2 + 2y + 1 = 1 - y^2
2y^2 + 2y = 0
2y(y+1) = 0
y = 0 or -1.

If y = 0, then we need cos x = 0 and sin x = 1, which occurs when x = pi/2 (plus multiples of 2pi).

If y = -1, then we need cos x = -1 and sin x = 0, which occurs when x = pi (plus multiples of 2pi).

Answer 5

cos x + 1 = sinx
x=π/2
0+1=0
x=π
-1+1=0

Answer 6

take sin x to the left and 1 to right

cos x - sinx = -1

we now cos (A+B) = cos A cos B - sin A sin B

take coefficient of cos x = 1 = A cos t
coeffiecient of sin x = 1 = A sin t
from above A = sqrt(2)
tan t = 1
t = pi/4
we get

cos x - sin x = sqrt(2)( cos t cos x - sin t sin x)
= sqrt(2)( cos (t+x)
= sqrt(2)cos( x + pi/4) = -1
cos (x + pi/4) = -1/sqrt(2) = cos (pi+pi/4)
so x + pi/4 = pi + pi/4 or -pi-pi/4
x = pi or pi/2

Answer 7

Plot the equations on separate sheets of tracing paper. Put the sheet with cos x vs x on top of the plot of sin x vs x, and slide it up one unit. Where the curves cross, the equation is solved.

Answer 8

cosx+1=sinx
cosx-sinx=-1
squaring both sides
cos^2x+sin^2x-2sinxcosx=1
1-sin2x=1
sin2x=0
2x=n*pi
x=n*pi/2