Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only.

1. sin2Į = (sqrt 3)/2
2. sin^2Į = cos^2Į + 1/2
3. sin 2x - cosx = 0
4. cos 4x = sin 2x


(((* stands for degrees)))

q1
sin2Į = (sqrt 3)/2
2Į = sin^-1 (sqrt3/2)
= 60*, 120*, 420*, 480*
Į = 30*, 60*, 210* and 240*

(not sure of the rest)

q2
sin^2Į = cos^2Į + 1/2
sin^2 Į - cos^2 Į = 1/2 ...(1)

sin^2 Į + cos^2 Į = 1 ...(2)

(1) + (2),
2sin^2 Į = 3/2
sin^2 Į = 3/4
sin ƒÆ = +-�ã3/2
Į = 60*, 120*, 240* and 300*

q3
sin 2x - cosx = 0
2sinxcosx - cosx = 0
cos x (2sin x - 1) = 0
cos x = 0 or sin x = 1/2
x = 90*, 270* or
x = 30* 150*

answer, x = 30*, 90*, 150* and 270*

q4
cos 4x = sin 2x
2(sin 2x)^2 + sin 2x - 1 = 0
(2sin 2x - 1)(sin 2x + 1) = 0
sin 2x = -1, 1/2
2x = 30*, 150*, 270*, 390*, 510*, 630*
x = 15*, 75*, 135*, 195*, 255* and 315*

""Į "" is supposed to be theta.

q1 - correct

q2 is correct but here is a simpler way.
Let Ø = ƒÆ
sin^2 Ø = cos^2 Ø + 1/2
cos^2 Ø - sin^2 Ø = -1/2
cos 2Ø = -1/2
2Ø = 120° , 240° ...
Ø = 60° , 120° , 240° , 300°

q3 correct

q4 also correct

good job!

To solve trigonometric equations, we need to use the properties of trigonometric functions and their inverse functions. Here's how you can solve each equation:

1. sin2Į = (sqrt 3)/2
To find the values of Į, we can take the inverse sine or arcsin of both sides:
2Į = arcsin((sqrt 3)/2)
2Į = 60 degrees or 120 degrees (using special angles)
Therefore, Į = 30 degrees or 60 degrees (positive values between 0 and 360 degrees).

2. sin^2Į = cos^2Į + 1/2
Using the identity sin^2Į + cos^2Į = 1, we can rewrite the equation as:
1 - cos^2Į = cos^2Į + 1/2
2cos^2Į = 1/2
cos^2Į = 1/4
cosƒÆ = ±1/2 (taking the square root)
Using inverse cosine or arccos, we find:
Į = 60 degrees, 120 degrees, 240 degrees, or 300 degrees (positive values between 0 and 360 degrees).

3. sin 2x - cosx = 0
Rearranging the equation:
2sinxcosx - cosx = 0
cosx(2sinx - 1) = 0
We have two cases to consider:
Case 1: cosx = 0
In this case, x = 90 degrees or x = 270 degrees.

Case 2: 2sinx - 1 = 0
Solving for sinx:
2sinx = 1
sinx = 1/2
Using inverse sine or arcsin:
x = 30 degrees or x = 150 degrees.
Therefore, x = 30 degrees, 90 degrees, 150 degrees, or 270 degrees (positive values between 0 and 360 degrees).

4. cos 4x = sin 2x
We can rewrite the equation using double-angle identities:
2(sin 2x)^2 + sin 2x - 1 = 0
Let's introduce a substitution: sin 2x = t.
2t^2 + t - 1 = 0
Factorizing:
(2t - 1)(t + 1) = 0
We have two cases to consider:
Case 1: 2t - 1 = 0
t = 1/2
Substituting back, we have:
sin 2x = 1/2
Using inverse sine or arcsin:
2x = 30 degrees, 150 degrees, 270 degrees, or 390 degrees
x = 15 degrees, 75 degrees, 135 degrees, or 195 degrees.

Case 2: t + 1 = 0
t = -1
Substituting back, we have:
sin 2x = -1
Using inverse sine or arcsin:
2x = 270 degrees or 630 degrees
x = 135 degrees or 315 degrees.
Therefore, x = 15 degrees, 75 degrees, 135 degrees, 195 degrees, 255 degrees, or 315 degrees (positive values between 0 and 360 degrees).

These are the positive values of the angles between 0 degrees and 360 degrees that satisfy each trigonometric equation.