Trig: how would you solve, for 0 degrees< A <90 degrees, 5sin2AsecA=3??

Question

I've tried substituting secA for 1/cosA but not really getting anywhere. Answer is supposed to be 17.5, but I have no idea how to get there! Any help would be much appreciated.

Answer 1

First sec A = 1/cos A
Next sin 2A = 2 sin A cos A
Since cos A is not 0 in the given range the equation becomes
5*2 sin A cos A/cos A = 10 sin A = 3.
So A = arcsin(3/10) = 17.5 degrees (approx.).

Answer 2

5sin(2A)secA = 3

substitute sin(2A) = 2sinAcosA and secA = 1/cosA

5*2sinAcosA(1/cosA) = 3

10sinA = 3

sinA = 3/10

A = sin^-1(3/10)

A = 12 pi/125 or 17.46 degrees

Answer 3

for 0 degrees< A <90 degrees, 5sin2AsecA=3
secA = 1/cosA
sin2A = 2 sinAcosA
so
5 sin2A secA = 5 sin2A/cosA
3= 5 (2sinAcosA)/cosA
10 sinA = 3
sinA = 0.3
A = 17.46 ~ 17.5 deg

Answer 4

5sin(2A)sec(A) = 3
5(2sin(A)cos(A)) / cos(A) = 3
10sinA = 3
sinA=3/10
A= 17.45760312°≈17.5°

A=17.5° ,

Answer 5

3sin(6x) * csc(2x) = 4 3sin(6x) / sin(2x) = 4 3sin(6x) = 4sin(2x) 3 * 2 * sin(3x) * cos(3x) = 4 * 2 * sin(x) * cos(x) 3 * (sin(2x)cos(x) + sin(x)cos(2x)) * (cos(2x)cos(x) - sin(2x)sin(x)) = 4 * sin(x) * cos(x) 3 * (2sin(x)cos(x)^2 + sin(x)cos(x)^2 - sin(x)^3) * (cos(x) * (cos(x)^2 - sin(x)^2) - 2sin(x)^2 * cos(x)) = 4 * sin(x) * cos(x) 3 * (3sin(x)cos(x)^2 - sin(x)^3) * (cos(x)^3 - sin(x)^2 * cos(x) - 2sin(x)^2 * cos(x)) = 4sin(x) * cos(x) 3 * (3sin(x) * (a million - sin(x)^2) - sin(x)^3) * (cos(x)^3 - 3sin(x)^2 * cos(x)) = 4sin(x) * cos(x) 3 * sin(x) * cos(x) * (3 * (a million - sin(x)^2) - sin(x)^2) * (cos(x)^2 - 3 * (a million - cos(x)^2)) = 4sin(x)cos(x) 3sin(x)cos(x) * (3 - 4sin(x)^2) * (4cos(x)^2 - 3) = 4sin(x)cos(x) 3sin(x)cos(x) * (12cos(x)^2 - 9 - 16sin(x)^2 * cos(x)^2 + 12sin(x)^2) = 4sin(x)cos(x) 3sin(x)cos(x) * (12 - 12sin(x)^2 - 9 + 12sin(x)^2 - 16sin(x)^2 * (a million - sin(x)^2) = 4sin(x)cos(x) 3sin(x)cos(x) * (3 - 16sin(x)^2 + 16sin(x)^4) = 4sin(x)cos(x) 3sin(x)cos(x) * (3 - 16sin(x)^2 + 16sin(x)^4) - 4sin(x)cos(x) = 0 sin(x)cos(x)* (9 - 48sin(x)^2 + 48sin(x)^4 - 4) = 0 sin(x) * cos(x) * (48sin(x)^4 - 48sin(x)^2 + 5) = 0 sin(x) = 0 x = 0 cos(x) = 0 x = ninety 48sin(x)^4 - 48sin(x)^2 + 5 = 0 sin(x)^2 = (forty 8 +/- sqrt(forty 8^2 - 4 * 5 * forty 8)) / (2 * forty 8) sin(x)^2 = (forty 8 +/- sqrt(forty 8 * forty 8 - 20 * forty 8)) / (ninety six) sin(x)^2 = (forty 8 +/- sqrt(28 * forty 8)) / ninety six sin(x)^2 = (forty 8 +/- sqrt(4 * 7 * 3 * sixteen)) / ninety six sin(x)^2 = (forty 8 +/- 8 * sqrt(21)) / ninety six sin(x)^2 = (6 +/- sqrt(21)) / 12 sin(x) = +/- sqrt( (6 +/- sqrt(21)) / 12) besides the incontrovertible fact that, because of the fact we are in quadrant a million, sin(x) could be advantageous sin(x) = sqrt( (6 +/- sqrt(21)) / 12 ) x = arcsin(sqrt( (6 +/- sqrt(21)) / 12 ))