We know that sin(3π/4) = (√2)/2, so we can rewrite the equation as:
sin(x + 3π/4) = -1/2
Now, we can determine the value of x by considering the unit circle. We know that the sin function is negative in the third and fourth quadrants, and its value is -1/2 at π/6 and 5π/6. Therefore, x is either π/6 or 5π/6.
Therefore, the solution to the equation sin(π/2+x)-cos(π+x) = 1 in the 10th grade is x = π/6 or x = 5π/6.
To start, we can use the sum-to-product trigonometric identity which states that sin(a) - cos(b) = 2sin((a+b)/2)sin((a-b)/2).
Therefore, we can rewrite the equation as:
2sin((π/2 + x + π + x)/2)sin((π/2 + x - π - x)/2) = 1
2sin((3π/2 + 2x)/2)sin((-π)/2) = 1
2sin(3π/4 + x)sin(-π/2) = 1
2sin(3π/4 + x)(-1) = 1
-2*sin(3π/4 + x) = 1
sin(3π/4 + x) = -1/2
We know that sin(3π/4) = (√2)/2, so we can rewrite the equation as:
sin(x + 3π/4) = -1/2
Now, we can determine the value of x by considering the unit circle. We know that the sin function is negative in the third and fourth quadrants, and its value is -1/2 at π/6 and 5π/6. Therefore, x is either π/6 or 5π/6.
Therefore, the solution to the equation sin(π/2+x)-cos(π+x) = 1 in the 10th grade is x = π/6 or x = 5π/6.