Solve sin 2 x = 0.8193, when 0≤ x ≤ 360˚? What quadrant (s) is it in?

Question

Solve sin 2 x = 0.8193, when 0≤ x ≤ 360˚? What quadrant (s) is it in?

Answer 1

Solve sin 2x = 0.8193, when 0˚ ≤ x ≤ 360˚?

What quadrant (s) is it in?

sin 2x = 0.8193, when 0˚ ≤ x ≤ 360˚

In this case because 0˚ ≤ x ≤ 360˚ and our equation contains 2x we need to look at the domain 0˚ ≤ 2x ≤ 720˚ (double everything)

Also using ASTC (All Stations to Chicgo or Cincinnatti or All Ships To California or whatever) you learn that All (A) trig ratios are positive in the first quadrant, Sine (S) is positive in the second, Tangent (T) is positive in the third and Cosine (C) is positive in the 4th

Now here sin 2x = 0.8193 which is positive so 2x is either in the 1st or second quadrants

So first find an acute angle whose sine is 0.8193

By calculator this angle is 55° (nearest degree)

Now 2x is in the 1st OR the 2ND quadrant

So 2x = 55° or (180° - 55°) and that is just for 0° to 360°

we need to keep going around again ... ( to get as far as 720°)

So 2x = 55° or (180° - 55°) or (360° + 55°) or (540° - 55°)

ie x = 27.5°, 62.5°, 207.5°, 242.5°
Note that these all satisfy the condition 0˚ ≤ x ≤ 360˚

BUT also note that 2x has to be in the 1st and 2nd quadrants and x ends up in the 1st and 3rd quadrants in this case.
This is because in fact 0˚ ≤ 2x ≤ 180˚ and 360˚ ≤ 2x ≤ 540˚ will ensure that sin 2x is positive.
So 0˚ ≤ x ≤ 90˚ and 180˚ ≤ x ≤ 270˚

Addendum: Acute angle forms of angles

These are important if in the end you are going to get the correct angles. The acute angle forms are really not difficult if you remember the relationship between the trig ratios and the unit circle. ie the x coordinate is cosθ and the y coordinate is sinθ

Then the acute angle form is the angle measured from the nearest x -axis in the first quadrant θ is acute (and is the same as α where α is alweays acute) or θ = 360 + α = etc

In the second quadrant θ = 180 - α = 540 - α = etc

In the third quadrant θ = 180 + α = 540 + α = etc

In the 4th quadrant θ = 360 - α = 720 - α etc

Answer 2

Sin 2x is a positive number and sine is positive for angles 0 to 180 degrees, so x can be in quadrant I (0-90 degrees) or in quadrant III (180-270 degrees).

Answer 3

in truth sin x = 0 skill that x = 0, +/- one hundred and eighty, +/- 360 etc. those type of values do not lie in any quadrant because the both lie between the first and fourth quadrants or between the 2d and 0.33 ones. some human beings used to call the angles that are multiples of ninety ranges as a quadrant angles. because they're on the boundary of those quadrants,

Answer 4

sin2x=0.8193
find out the angle whose sin is 0.8193
equate it to 2x
find x by dividing by 2
that will dive one solution
for the other solution subtract the first solution from 180*