Find all solutions w between 0 and 360, inclusive:

(a) cosw = cos(−340)
(b) cosw = sin20
(c) sinw = cos(−10)
(d) sinw < − 1 2
(e) 1 < tanw

Let's solve each equation step-by-step:

(a) cosw = cos(−340):

The cosine function is a periodic function with a period of 360 degrees.

To find the solutions within the range of 0 to 360 degrees, we can use the fact that the cosine function is an even function, which means cos(-x) = cos(x).

Therefore, cosw = cos(-340) is equivalent to cosw = cos(340).

To find the angles that satisfy this equation, we can use the inverse cosine function:

w = arccos(cos(340))

Using a calculator, we find that arccos(cos(340)) = 20 degrees.

The solutions within the range of 0 to 360 degrees are w = 20 degrees and w = 340 degrees.

(b) cosw = sin20:

To find the solutions for this equation, we can rearrange it as cosw - sin20 = 0.

To solve this equation, we can express sin20 in terms of cosine:

sin20 = cos(90 - 20) = cos70.

Now, the equation becomes cosw - cos70 = 0.

Using the fact that the cosine function is an even function, we have cos70 = cos(-70), so the equation becomes cosw - cos(-70) = 0.

Again, we can solve this using the inverse cosine function:

w = arccos(cos(-70))

Using a calculator, we find that arccos(cos(-70)) = 70 degrees.

The solution within the range of 0 to 360 degrees is w = 70 degrees.

(c) sinw = cos(−10):

To find the solutions for this equation, we can rearrange it as sinw - cos(-10) = 0.

Since the sine function is an odd function, sin(-x) = -sin(x), we have sin(-10) = -sin(10).

Therefore, the equation becomes sinw + sin(10) = 0.

To solve this equation, we can use the inverse sine function:

w = arcsin(-sin(10))

Using a calculator, we find that arcsin(-sin(10)) = -10 degrees.

The solution within the range of 0 to 360 degrees is w = -10 degrees.

(d) sinw < -1/2:

To find the solutions for this inequality, we need to find the angles where sine is less than -1/2.

The possible solutions are in the fourth quadrant (180 to 270 degrees) and the third quadrant (270 to 360 degrees).

In the fourth quadrant, sinw is negative. Using the fact that sine is an odd function, we have sin(-x) = -sin(x). Therefore, sinw < -1/2 becomes -sinw > 1/2.

Using the inverse sine function, we have:

-w > arcsin(1/2)
-w > 30
w < -30

In the third quadrant, sinw is also negative. So we have:

w > 180 - arcsin(1/2)
w > 180 - 30
w > 150

Combining the two ranges, the solutions within the range of 0 to 360 degrees are w < -30 and w > 150.

(e) 1 < tanw:

To find the solutions for this inequality, we need to find the angles where the tangent function is greater than 1.

The possible solutions are in the first quadrant (0 to 90 degrees) and the third quadrant (270 to 360 degrees).

In the first quadrant, the tangent function is positive. Using the inverse tangent function, we have:

w > arctan(1)
w > 45

In the third quadrant, the tangent function is negative. Using the inverse tangent function, we have:

w < arctan(-1)
w < -45

Combining the two ranges, the solutions within the range of 0 to 360 degrees are w > 45 and w < -45.

To find all solutions w between 0 and 360 (inclusive) for the given trigonometric equations, you can use the properties and periodicity of trigonometric functions along with some algebraic manipulations. Here's how to solve each equation:

(a) cosw = cos(-340)
Since cosine is an even function, cos(-θ) = cos(θ). Therefore, we can rewrite the equation as:
cosw = cos(340)

To find the solutions, we can equate the angles within the same period:
w = 340

So, the solution to the equation is w = 340.

(b) cosw = sin20
Since cosine and sine functions are different, we need to approach this equation differently. We can rewrite the equation as:
cosw - sin20 = 0

To solve this equation, we need to use the identities involving sine and cosine. One such identity is:
sin²θ + cos²θ = 1

Rearranging the identity, we get:
cos²θ = 1 - sin²θ

Substituting this into the original equation, we have:
cosw = √(1 - sin²20)

Now, we can solve for cosw:
cosw = √(1 - sin²20)
cosw = √(1 - (1/2)²)
cosw = √(1 - 1/4)
cosw = √(3/4)
cosw = √3/2

The value √3/2 represents the cosine of an angle that is 30 degrees. However, since we are looking for all solutions between 0 and 360, we need to consider both the positive and negative 30 degrees.

So, the solutions to the equation are w = ±30.

(c) sinw = cos(-10)
Using the same logic as in part (a), we can rewrite the equation as:
sinw = cos(10)

To find the solutions, we can equate the angles within the same period:
w = 10

So, the solution to the equation is w = 10.

(d) sinw < -1/2
To solve this inequality, we need to find all angles within the specified range that have a sine value less than -1/2.

First, recall the unit circle to visualize the sine values for different angles. The value -1/2 corresponds to an angle of -30 degrees or 210 degrees on the unit circle.

Since we want the negative sine values, we consider the angles in the fourth quadrant (180 to 270 degrees) and those in the third quadrant (270 to 360 degrees). These angles will have negative sine values.

The solutions are: 210 and 330 degrees.

(e) 1 < tanw
To solve this inequality, we need to find all angles within the specified range that have a tangent value greater than 1.

Recall that the tangent function is positive in the first and third quadrants. Specifically, for any angle θ, tanθ is greater than 1 if it lies between 45 and 135 degrees or between 225 and 315 degrees.

So, the solutions are: 45, 135, 225, 315 degrees.