12. Find the values in the interval [-n/2, n] that satisfy the equation.
arctan (V3/3) = x
a. since tan (n/6) = V3/3, X = n/6
b. since sin (-1/4)= -V2/2, x = -1/4
c. since cos (3n/2) = -1/2, X = 3n/2
d. since cos 0 = 1, x = 0
e. since cos (n/4) = V2/2, X = n/4
13.
[tex]cot^2x = 4/12 \\ cot^2x = 1/3 \\ tan^2x = 3 \\ tan(x)= +- \sqrt{3} [/tex]
so, [tex]x= \pi/3 \\ x = \pi - \pi/3 = 2\pi/3 \\ x = \pi + \pi/3 = 4\pi/3 \\ x = 2\pi - \pi/3 = 5\pi/3 [/tex]
x ∈ {2π/3, π, 4π/3} ≈ {2.09440, 3.14159, 4.18879}
Step-by-step explanation:
The equation can be put into standard form by adding 1:
2cos²(x) +3cos(x) +1 = 0
(2cos(x) +1)(cos(x) +1) = 0
Values of cos(x) that make this true* are ...
cos(x) = -1/2 . . . . . . . . . true for x=2π/3, x=4π/3
cos(x) = -1 . . . . . . . . . . . true for x=π
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A graphing calculator can be helpful here, too.
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* from your knowledge of the short table of trig functions and their signs in different quadrants
[tex]Find all real numbers on the interval [0.2 t) that satisfy the equation. Use radian measure.2 cos (e[/tex]
[tex]Find all real numbers on the interval [0.2 t) that satisfy the equation. Use radian measure.2 cos (e[/tex]
The value of t that will satisfy the equation is π/6 (which is 30 degrees)
Step-by-step explanation:
The function that models the movement of the particle is given as;
S(t) = 2 sin(t) + 3 cos (t)
Now we want to the value of t between 0 and pi/2 that satisfies the equation;
s(t) = (2+ 3√3)/2 = 1 + 3√3/2
What we do here is simply find that value of t that would ensure that;
2sin(t) + 3cos(t) = 1 + 3√3/2
Without any need for rigorous calculations, this value of t can be gotten by inspection.
From our regular trigonometry, we know that the sin of angle 30 is 1/2 and its cos value is √3/2
We can make a substitution for it in this equation.
We obtain the following;
2 sin(30) + 3cos (30) and that is exactly equal to 1 + 3√3/2
Do not forget however that we have a range. And the range in question is between 0 and π/2
Kindly that π/2 in degrees is 90 degrees
So our range of values here is between 0 and 90 degrees.
So to follow the notation in the question, the value within the range that will satisfy the equation is π/6
(0,-5)(1,-6) is the answers
6 sin 2x = 6 cos x Using the identity sin 2x = 2 sin x cos x:-
12 sin x cos x = 6 cos x
6 cos x ( 2 sin x - 1) = 0
either 6 cos x = 0 or 2 sin x - 1 = 0 so sin x = 1/2
x = pi/2, 3pi/2 , pi/6, 5pi / 6
answer is ( pi/6, pi/2, 5pi/6, 2pi/2)
[tex]\sin(2x) = \cos(x) \\ 2 \sin(x) \cos(x) = \cos(x) \\ \cos(x) (2 \sin(x) - 1) = 0[/tex]
so
[tex]\cos(x) = 0 x = \frac{\pi}{2} + k\pi \\ \sin(x) = \frac{1}{2} x = \frac{\pi}{6} + 2k\pi \\ or \: x = \frac{5\pi}{6} + 2k\pi[/tex]
[tex]x = \frac{\pi}{2} or \frac{3\pi}{2} \\ x = \frac{\pi}{6} or \frac{5\pi}{6}[/tex]
The solutions are π/4, 3π/4,5π/4,7π/4
Step-by-step explanation:
The given equation is
6sin²(x) = 3
Divide by 6 to get:
[tex]{ \sin}^{2} (x) = \frac{1}{2}[/tex]
This implies that;
[tex]\sin(x) = \pm \frac{ \sqrt{2} }{2}[/tex]
If
[tex]\sin(x) = \frac{ \sqrt{2} }{2}[/tex]
[tex]x = \frac{\pi}{4}[/tex]
in the first quadrant
[tex]x = \frac{3\pi}{4}[/tex]
in the second quadrant.
If
[tex]\sin(x) = - \frac{ \sqrt{2} }{2}[/tex]
[tex]x = \frac{5\pi}{4}[/tex]
in the third quadrant
[tex]x = \frac{7\pi}{4}[/tex]
3 sin ( 2 x ) = 3 cos x
3 · 2 sin x cos x - 3 cos x = 0
6 sin x cos x - 3 cos x = 0 /: 3
2 sin x cos x - cos x = 0
cos x ( 2 sin x - 1 )
cos x = 0,
x 1 = π/2, x 2 = 3π/2;
2 sin x - 1 = 0
2 sin x = 1
sin x = 1/2
x 3 = π/6, x 4 = 5π/6.
The given equation is
[tex]8cos(x)+ 4 sin(2x)=0[/tex]
Using the double angle formula of sin(2x),
[tex]8cos(x) +4(2) sin(x)cos(x)=0[/tex]
[tex]8cos(x)+ 8 sin(x) cos(x)=0[/tex]
Taking out 8 cos x
[tex]8cos(x) (1+ sin(x)) =0[/tex]
[tex]8 cos(x)=0 , 1+sin (x)=0 \\ 8cos (x)=0, sin(x)=-1 \\ cos(x)=0 , sin(x)=-1 \\ x = \frac{ \pi}{2} , \frac{ 3 \pi}{2}[/tex]
And that's the required answer.
Step-by-step explanation:
[tex]tan^{-1} (\frac{\sqrt{3} }{3} )=x\\tan x=\frac{\sqrt{3} }{3} =tan (\frac{\pi}{6} )\\x=\frac{\pi}{6}[/tex]