## 1) Solve the equation in the interval [0Â°, 360Â°]:

a) sin 2x = -sin x:

To solve this equation, we can use the double angle formula for sine, which states that sin 2x = 2sin x cos x.

So, we have 2sin x cos x = -sin x.

Dividing both sides of the equation by sin x, we get:

2cos x = -1.

Now, we need to solve for x. To do that, let's consider the possible values for cos x when it is equal to -1/2.

Using the unit circle or the trigonometric ratios, we find that cos x = -1/2 when x = 120Â° and x = 240Â°.

However, we have to check the validity of these solutions by substituting them back into the original equation. We see that x = 120Â° satisfies the equation, while x = 240Â° does not.

Therefore, the solution to the equation sin 2x = -sin x in the interval [0Â°, 360Â°] is x = 120Â°.

b) sin 2T = -1/2:

Similar to part a, we can use the double angle formula for sine to rewrite the equation as 2sin T cos T = -1/2.

Dividing both sides by 2 and rearranging, we get:

sin T cos T = -1/4.

Now, we need to solve for 2T. We can find the value for 2T that satisfies this equation by considering the values of sin T and cos T when their product is equal to -1/4.

Using the unit circle or the trigonometric ratios, we find that sin T = -1/2 and cos T = -1/2 satisfy the equation.

Now, solve for 2T:

2T = sin^-1(-1/2) = 210Â° or 330Â°.

Finally, divide 2T by 2 to obtain the solutions for T: T = 105Â° or 165Â°.

Therefore, the solutions to the equation sin 2T = -1/2 in the interval [0Â°, 360Â°] are T = 105Â° and T = 165Â°.

c) 4 sin^2T = 3:

First, let's isolate sin^2T by dividing both sides of the equation by 4:

sin^2T = 3/4.

Next, take the square root of both sides to find sin T:

sin T = Â±âˆš(3/4) = Â±âˆš3/2.

Since sin T can be positive or negative, we need to consider both possibilities.

Using the unit circle or the trigonometric ratios, we find that sin T = âˆš3/2 when T = 60Â° or 300Â°.

Similarly, sin T = -âˆš3/2 when T = 120Â° or 240Â°.

Therefore, the solutions to the equation 4 sin^2T = 3 in the interval [0Â°, 360Â°] are T = 60Â°, T = 120Â°, T = 240Â°, and T = 300Â°.

2) Evaluate the expression:

sin(arctan 2):

To evaluate this expression, first find the value of arctan 2. This can be done using a calculator or by finding the angle whose tangent is 2. The value of arctan 2 is approximately 63.43Â°.

Now, we need to find the sine of 63.43Â°. Using either a calculator or the unit circle, we find that sin(63.43Â°) is approximately 0.896.

Therefore, sin(arctan 2) â‰ˆ 0.896.

3) Rewrite the following without using trigonometric or inverse trigonometric functions:

cos(arcsin u):

When arcsin u = Î¸, it means that sin Î¸ = u.

Let's assume that sin Î¸ = u and cos Î¸ = v.

Now, using basic trigonometric properties:

sin^2 Î¸ + cos^2 Î¸ = 1.

Substitute the values of sin Î¸ and cos Î¸:

u^2 + v^2 = 1.

Since sin Î¸ = u, we have:

v = Â±âˆš(1 - u^2).

Therefore, cos(arcsin u) can be rewritten as Â±âˆš(1 - u^2) depending on the positive or negative value of v.