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.
a.)
( I got confused doing this 1 can you help me with it.)
3 sin theta - 4 cos theta = 2
b.)
(can you correct this 1 for me.)
tan (x+15)= 3 tan x
tan(x+15) = (tanx + tan15)/(1-tanx tan15)
(tanx + tan15)/(1-tanx tan15) = 3 tanx
tanx + tan 15 = 3 tanx - 3tan2x tan15
3 tan2x tan15 - 2tanx + tan15 = 0
tan15 = 0.27, denote tanx as t
0.81 t2 - 2t + 0.27 = 0
D = 0.1252
t1 = 1.67
t2 = 0.80
tan (x1) = 1.67
x1 = arctan (1.67) + ¥ðn = 59.087 + 180n = 59¨¬ 5'+ 180n, n=0,1,2,..
tan(x2) = 0.80
x2 = arctan(0.80) + ¥ðn = 38.66 + 180n = 38¨¬ 40' + 180n, n=0,1,2,..
Answer: 59¨¬ 5', 239¨¬ 5', 38¨¬ 40', 218¨¬ 40'
Answer for b. 29deg 5',239deg 5',38deg 40',218deg 40'.
can you plz check if i messed anything.
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a.) To solve the equation 3 sin(theta) - 4 cos(theta) = 2, we can use the identities sin(theta) = cos(90 - theta) and cos(theta) = sin(90 - theta) to rewrite the equation in terms of a single trigonometric function.
Let's use sin(theta) = cos(90 - theta) to rewrite the equation:
3 cos(90 - theta) - 4 cos(theta) = 2
Applying the distributive property, we get:
3 cos(90) cos(theta) + 3 sin(90) sin(theta) - 4 cos(theta) = 2
Simplifying further:
3 sin(theta) - 4 cos(theta) = 2
This is the original equation, so we haven't made progress yet. We need to try a different approach. One way to solve this equation is by using the double-angle formula for sine.
The double-angle formula for sine is sin(2theta) = 2sin(theta)cos(theta)
Let's rewrite the equation using sin(2theta):
3 sin(theta) - 4 cos(theta) = 2
6 sin(theta) cos(theta) - 4 cos(theta) = 2
2 cos(theta) (3 sin(theta) - 2) = 2
Now, we have a product of two factors equal to 2. For this equation to be true, either one or both of the factors must be equal to 2.
Setting 2 cos(theta) = 2:
cos(theta) = 1
theta = arccos(1) = 0 degrees
Setting 3 sin(theta) - 2 = 2:
3 sin(theta) = 4
sin(theta) = 4/3
theta = arcsin(4/3) ≈ 63.4 degrees
So, the positive values of theta between 0 degrees and 360 degrees that satisfy the equation are approximately 0 degrees and 63.4 degrees.
b.) The given equation is tan(x + 15) = 3 tan(x).
First, we can simplify the equation using the tangent addition formula:
tan(x + 15) = (tan(x) + tan(15))/(1 - tan(x) tan(15))
Now, substitute this into the equation:
(tan(x) + tan(15))/(1 - tan(x) tan(15)) = 3 tan(x)
Next, let's simplify the equation by multiplying both sides by 1 - tan(x) tan(15):
(tan(x) + tan(15)) = 3 tan(x)(1 - tan(x) tan(15))
Expand the right side of the equation:
tan(x) + tan(15) = 3 tan(x) - 3 tan^2(x) tan(15)
Rearrange the terms:
3 tan^2(x) tan(15) - 2 tan(x) + tan(15) = 0
Now, we can substitute tan(15) with its approximate value, 0.27:
3 tan^2(x) (0.27) - 2 tan(x) + 0.27 = 0
Next, let's solve this quadratic equation for tan(x). Using the quadratic formula, we have:
tan(x) = (-(-2) ± sqrt((-2)^2 - 4(3)(0.27)))/(2(3))
tan(x) = (2 ± sqrt(4 - 3.24))/6
tan(x) = (2 ± sqrt(0.76))/6
Now, we can approximate the values of tan(x):
tan(x1) ≈ (2 + sqrt(0.76))/6 ≈ 1.67
tan(x2) ≈ (2 - sqrt(0.76))/6 ≈ 0.80
To find the angles x1 and x2, we can use the arctangent function (atan):
x1 = atan(1.67) + 180n ≈ 59.087 + 180n ≈ 59° 5' + 180n, where n = 0, 1, 2, ...
x2 = atan(0.80) + 180n ≈ 38.66 + 180n ≈ 38° 40' + 180n, where n = 0, 1, 2, ...
Therefore, the positive values of x between 0 degrees and 360 degrees that satisfy the equation are approximately 59° 5', 239° 5', 38° 40', and 218° 40'.