(a) 3 cos (3x + 2) = 2 over the interval (-π, π]. (b) 2 sin(2x - 3) = 4 over the interval (0, 2. π]. (c) 2 cos(2x - 3) = 0 over the interval (-, 3. π]. Pay attention to the interval. (d) tan(2x - 3) = 2 over the interval (-3, 3-π].

Q: -p² sin(w) cos?(w) – p cos² (») -p2 sin(4).

A:

Q: θ = cos^-1[sin(LT1) sin(LT2) + cos(LT1) cos(LT2) cos(LN1 − LN2)] How does this formula go with…

A: Consider the figure as our Earth. O is the center of the earth Point P on the earth surface has…

Q: (b) S tan3 x sec² x dx c) S sin³x cos5 x dx

A: The explanation of the solution is :

Q: 4 sec(8) = find cot(0) 2

A: Use trigonometric rations

Q: 2. Explain why arcsin(sin(0)) = 0 but arcsin(sin(807)) # 807

A: Concept  arcsin(sin x) = x   is true  only when x lie in the interval [-π/2 , π/2]

Q: First, recall that sin²2 (0) = 1 - cos² (0). Thus, substitute (1 - cos²(0)) for sin²(0) in the given…

A: Now we know that sin^2(theta)=1-cos^2(theta)

Q: 6) y=-sin (3x) - 1

A: 6. The given function y=-sin3πx-1. We have to sketch the graph of the function and find its domain…

Q: 8.) k(x)=-2 tan(2(x+)) Period = Increment =

A: Given function is: kx=-2tan2x+π8 Then the period of this function is 2π2=π. The increment is π8.

Q: Wouldn't tan(x+pi)-(-sin(x)) yield tan(x+pi) + sin(x)) instead of tan(x+pi) - sin(x))?

A: Yes , there is a mistake on that question. I give u correct solution .  tan(x + π) - cos( x + (π/2))…

Q: (cos A + sin A)/(cos A - sin A) = (1+ sin(2A))/ cos(2A)

A:

Q: cos(4x) cos(2x) sin(4x) sec(2x) 12. Verify: - sin(2x)

A: cos(2x ) =cos2(x)-sin2(x) = 2cos2(x)-1 =1-2sin2(x) sin(2x) = 2sin(x) cos(x)