a.)Find sin 4x if tan=0.6

b.) Solve: sinxsinx/2=1-cosx(0<x<2*pie

a.) To find sin 4x, we'll use the given information that tan is equal to 0.6. Let's break down the steps to get the answer:

Step 1: Start with the given information that tan = 0.6. We know that tan is equal to sin/cos, so we can write:

tan = sin/cos

Step 2: Rearrange the equation to isolate sin:

sin = tan * cos

Step 3: Find cos. To do this, we can make use of the Pythagorean Identity, which states:

sin^2 + cos^2 = 1

Given that sin is squared in the equation, we can rewrite the equation as:

1 - sin^2 = cos^2

Step 4: Solve for cos:

cos = √(1 - sin^2)

Step 5: Substitute the value of tan from the given information into the equation to get sin:

sin = tan * √(1 - sin^2)

Now we have an equation for sin in terms of the given tan value.

b.) To solve the equation sin(x)*sin(x/2) = 1 - cos(x) for the interval 0 < x < 2π, we'll go through the following steps:

Step 1: Simplify the equation:

sin(x)*sin(x/2) = 1 - cos(x)

Step 2: Use the double-angle identity for sine, which states:

sin(2θ) = 2*sin(θ)*cos(θ)

We can rewrite the equation as:

2*sin(x/2)*cos(x/2)*sin(x/2) = 1 - cos(x)

Step 3: Simplify further:

2*sin^2(x/2)*cos(x/2) = 1 - cos(x)

Step 4: Apply the Pythagorean Identity to eliminate the cosine terms:

2*(1 - cos^2(x/2))*cos(x/2) = 1 - cos(x)

Step 5: Simplify and rearrange the equation:

2*cos(x/2) - 2*cos^3(x/2) = 1 - cos(x)

Step 6: Use the identity: cos^2(θ) = (1 + cos(2θ))/2

We can rewrite the equation as:

2*cos(x/2) - 2*((1 + cos(x))/2)^3 = 1 - cos(x)

Step 7: Simplify the equation further:

2*cos(x/2) - (1 + cos(x))^3 = 1 - cos(x)

Step 8: Expand and simplify:

2*cos(x/2) - (1 + 3*cos(x) + 3*cos^2(x) + cos^3(x)) = 1 - cos(x)

Step 9: Group like terms and arrange the equation:

2*cos(x/2) - 1 - 3*cos(x) - 3*cos^2(x) - cos^3(x) = 1 - cos(x)

Step 10: Combine like terms and eliminate the common factor:

2*cos(x/2) - 1 - 4*cos(x) - 3*cos^2(x) - cos^3(x) = 0

Now we have a simplified equation that we can solve for the given interval.