Question

a. 0b. pi/2c. pid. 3pi/2These are 4 options but there can be more than 2 or 3 correct answers.Find the solution of each equation the interval( 0, 2pi).

Answer 1

0 and pi

Step-by-step explanation:

Given the expression

sinx - cosx = 1

Square both sides of the equation

(sinx - cosx)^2= 1^2

sin^2x -2sinxcosx + cos^2x = 1

sin^2x +cos^2x - 2sinxcosx = 1

In trigonometry identity, sin^2x +cos^2x = 1

The expression becomes;

1 - 2sinxcosx = 1

-2sinxcosx = 1-1

-2sinxcosx = 0

2sinxcosx = 0

sinxcosx = 0/2

sinxcosx = 0

Also sinxcosx = sin2x/2

Substitute in the result, we will have:

sin2x/2 = 0

sin2x = 2(0)

sin2x = 0

2x = arcsin0

2x = 0

x = 0/2

x = 0

Since sin is positive in the second quadrant, x = 180 + 0

x = 180 degrees or pi rad

Hence the value of x within the interva; (0, 2pi) are 0 and pi