Solve Tan^2x/2-2 cos x = 1 for 0 < or equal to theta < greater or equal to 1.
1 answer:
Answer:
x = theta = 0°
Step-by-step explanation:
Given the trigonometry function
Tan²x/2-2 cos x = 1
Tan²x-4cosx = 2 ... 1
From trigonometry identity
Sec²x = tan²x+1
tan²x = sec²x-1 ... 2
Substituting 2 into 1, we have:
sec²x-1 -4cosx = 2
Note that secx = 1/cosx
1/cos²x - 1 - 4cosx = 2
Let cosx. = P
1/P² - 1 - 4P = 2
1-P²-4P³ = 2P²
4P³+2P²+P²-1 = 0
4P³+3P² = 1
P²(4P+3) = 1
P² = 1 and 4P+3 = 1
P = ±1 and P = -3/4
Since cosx = P
If P = 1
Cosx = 1
x = arccos1
x = 0°
If x = -1
cosx = -1
x = arccos(-1)
x = 180°
Since our angle must be between 0 and 1 therefore x = 0°
You might be interested in
Answer:62
Step-by-step explanation:Solution for What is 20 percent of 310:
20 percent *310 =
(20:100)*310 =
(20*310):100 =
6200:100 = 62
Answer:
220 I'm pretty sure because 6×5 is 30 times 2 is 60. Then do 6×10 equals 60. Then do 5 times 10 times 2 to get 100. Then you add it all up to get 220
12/3 = 4
it was dilated with a scale factor of 4
For sake of convenience, reverse the series, then it would be:
21,19,17.....7
a=21, d = -2,
s(4) = 4/2 (21+15)
s(4)=2(36)
s(4) = 72