tan(3x)=1
Let p=3x
Tanp=1
P=tan-¹(1)
P=π/4=45°
Where
P=3x
Tan has a period of π
3x=π/4+πn
X=π/12+πn/3
Where n represent integer
Similar
3sec^2x=4
3/cos²x=4
Cos²x=3/4
Cosx=√3/2=30° or π/6
Cos has a period movement of 2π
x=π/6+2nπ
What ever you do make sure you don't exceed 2π
If you fancy degree (π=180)
Find the solution to the follow trig equations on the interval [0,2pi). Round to the nearest ten thousandth. If there are multiple values, enter your answers from least to greatest separated by commas without spaces. Example: .1234,.5678,.9101
1) tan(3x)=1
2) 3sec^2x=4
3 answers
I find these are easiest done this way:
tan(3x)=1
I know that tan 45° =1 or tan225° = 1, (using the CAST rule)
3x = 45° or 3x = 225°
x = 15° or x = 75°
but the period of tan(3x) is 360/3 = 120°
so to find a new solution all we have to do is add 120° to any current solution
to stay within the 0 ≤ x ≤ 2π domain
x = 15°, 135°, 255°, 75°, 195°, or 315°
or in radians, π/12, 9π/12, 17π/12, 13π/12, 21π/12
3sec^2x=4
sec^2x=4/3
secx = ± 2/√3 , so x can be in all 4 quadrants
cosx = √3/2 , recognize x = 30°
x = 30°, 150°, 210°, or 330°
or the equivalent radians of π/6, 5π/6, 7π/6 and 11π/6
tan(3x)=1
I know that tan 45° =1 or tan225° = 1, (using the CAST rule)
3x = 45° or 3x = 225°
x = 15° or x = 75°
but the period of tan(3x) is 360/3 = 120°
so to find a new solution all we have to do is add 120° to any current solution
to stay within the 0 ≤ x ≤ 2π domain
x = 15°, 135°, 255°, 75°, 195°, or 315°
or in radians, π/12, 9π/12, 17π/12, 13π/12, 21π/12
3sec^2x=4
sec^2x=4/3
secx = ± 2/√3 , so x can be in all 4 quadrants
cosx = √3/2 , recognize x = 30°
x = 30°, 150°, 210°, or 330°
or the equivalent radians of π/6, 5π/6, 7π/6 and 11π/6
Cool I like the way you think smart .......you make maths fun to learn here