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secondary 3 | A Maths
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Date Posted: 4 years ago
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Condition : 0° < x < 360°
sin x - sin x tan x = 0
Question tells you a hint which is take out the common factors. The common factor here is sin x .
So, sin x(1 - tan x) = 0
sin x = 0 or 1 - tan x = 0
sin x = 0 or tanx = 1
For sinx = 0, basic angle = sin-¹(0) = 0°
Typically, x = 0°,180°,360°,540°.. and so on.
or x = 0°, -180°, -360°, -540°..
Every time we move 180° or π radians backward or forward from 0°/0, our sin x is 0.
But due to our condition above, only 180° satisfies the range.
So x = 180°
Now we look at tan x = 1
basic angle = tan-¹(1) = 45°
tan x is positive so x lies in the 1st and 3rd quadrant.
x = 45°, 45° + 180°
x = 45°, 225°
∴ x = 45°, 180°, 225°
sin x - sin x tan x = 0
Question tells you a hint which is take out the common factors. The common factor here is sin x .
So, sin x(1 - tan x) = 0
sin x = 0 or 1 - tan x = 0
sin x = 0 or tanx = 1
For sinx = 0, basic angle = sin-¹(0) = 0°
Typically, x = 0°,180°,360°,540°.. and so on.
or x = 0°, -180°, -360°, -540°..
Every time we move 180° or π radians backward or forward from 0°/0, our sin x is 0.
But due to our condition above, only 180° satisfies the range.
So x = 180°
Now we look at tan x = 1
basic angle = tan-¹(1) = 45°
tan x is positive so x lies in the 1st and 3rd quadrant.
x = 45°, 45° + 180°
x = 45°, 225°
∴ x = 45°, 180°, 225°
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