If sin(2x) = cos (x + 30%), what is the value of x?!
1 answer:
Answer:
x ∈ {20°, 120°, 140°, 260°}
Step-by-step explanation:
I find a graphing calculator useful for solving equations of this sort. Subtracting cos(x+30°) from both sides gives an expression that is zero when x has the appropriate value. The calculator is very good at finding zeros.
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We can make use of the formula for the sum of two sines.
sin(a) +sin(b) = 2sin((a+b)/2)cos((a-b)/2)
Then we can rewrite ...
sin(2x) -sin(x +120°) = 0
as ...
2sin((2x-(x+120°))/2)·cos((2x+(x+120°))/2) = 0
sin((x -120°)/2)·cos((3x +120°)/2) = 0 . . . . . divide by 2 and simplify a bit
This resolves to two equations according to the zero product rule.
x/2 -60° = k·180° . . . . . for some integer k
x = 120° +k·360° . . . . . for some integer k
And the other equation is ...
3x/2 +60° = (2k +1)·90° . . . . . for some integer k
3x/2 = k·180° +30° . . . . . . . . subtract 60°, eliminate parentheses
3x = k·360° +60°
x = k·120° +20° . . . . . for some integer k
The values shown in the Answer section above are the values of x in the range 0 to 360°: 20°, 120°, 140°, 260°.
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Answer:
(2,-5) and (2,1).
Step-by-step explanation:
We need to find the find all the points having an x coordinate of 2 whose distance from the point (-2,-4) is 5.
A circle with center (-2,-4) and radius 5 represents all the points whose distance from the point (-2,-4) is 5.
Standard form of a circle is
where, (h,k) is center and r is the radius.
Now, put x=2.
Taking square root on both sides.
So, the y-coordinates of the points are -5 and 1.
Therefore, the required points are (2,-5) and (2,1).
Answer:
0.000329
Step-by-step explanation:
Please have a look at attached document.you will get complete solution.
