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How do you solve #cosx-4=sinx-4# for #0<=x<=2pi#?

Answer 1

Aaliyah Abdullah

#x= pi/4, x=(5pi)/4# for #0<= x <= 2pi#

#cosx-4=sinx-4 or cosx= sinx or tanx=1 ; tan(pi/4)=1 ; tan(pi+pi/4)=1 or tan((5pi)/4)=1 :. x= pi/4, x=(5pi)/4#[Ans]

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Answer 2

Elijah Alexander

To solve the equation cos(x) - 4 = sin(x) - 4 for 0 <= x <= 2pi:

Subtract 4 from both sides of the equation: cos(x) - 4 - 4 = sin(x) - 4 - 4 cos(x) - 8 = sin(x) - 8
Since cos(x) - 8 = sin(x) - 8, we can add 8 to both sides: cos(x) - 8 + 8 = sin(x) - 8 + 8 cos(x) = sin(x)
Use the Pythagorean identity to rewrite sin(x) as sqrt(1 - cos^2(x)): cos(x) = sqrt(1 - cos^2(x))
Square both sides of the equation to eliminate the square root: (cos(x))^2 = (sqrt(1 - cos^2(x)))^2 cos^2(x) = 1 - cos^2(x)
Rearrange the equation: 2cos^2(x) = 1
Divide both sides by 2: cos^2(x) = 1/2
Take the square root of both sides: cos(x) = ±sqrt(1/2)
Since 0 <= x <= 2pi, the solutions are: x = π/4, 3π/4, 5π/4, and 7π/4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #cosx-4=sinx-4# for #0&lt;=x&lt;=2pi#?

How do you solve #cosx-4=sinx-4# for #0<=x<=2pi#?