The number of values in the interval –π ≤ x ≤ π that satisfy the equation sin x = 2 cos x is
1
2
3
4
0
sinx = 2cosx
tanx = 2
tanx > 0 in QI or QIII, so there are two values for x.
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To find the number of values in the given interval that satisfy the equation sin x = 2 cos x, we can use the trigonometric identity sin x = cos(π/2 - x).
Substituting this identity into the equation, we get cos(π/2 - x) = 2 cos x.
Now, let's simplify this equation. Using the cosine difference identity, cos(π/2 - x) = cos π/2 cos x + sin π/2 sin x = 0 + sin x = 2 cos x.
This simplifies to sin x = 2 cos x.
Since the given equation is equivalent to sin x = 2 cos x, we can say that the number of values in the interval –π ≤ x ≤ π that satisfy this equation is equal to the number of values of x within this interval that satisfy sin x = 2 cos x.
To find these values, we can use the fact that sin x = cos 90° - x.
Now, we have sin x = cos 90° - x = 2 cos x.
Taking the cosine of both sides, we get cos x = cos (90° - x).
Since the cosine function is an even function, meaning that cos(-x) = cos x, we can rewrite the equation as cos x = cos (x - 90°).
Now, we need to find the x values that satisfy cos x = cos (x - 90°).
For two angles to have the same cosine value, the angles must either be equal or one must be the negative of the other. So, we have two possibilities:
1) x = x - 90°. This equation simplifies to -90° = 0°, which is not possible.
2) x = -(x - 90°). This equation simplifies to 2x = 90°, which gives x = 45°.
Therefore, the only value of x within the interval –π ≤ x ≤ π that satisfies the equation sin x = 2 cos x is x = 45°.
So, the number of values in the given interval that satisfy the equation sin x = 2 cos x is 1.