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Zorluk: OrtaTrigonometric Graphs and Simple Equations

For the domain 0∘≤x≤180∘0^\circ \le x \le 180^\circ, what is the complete solution set to the trigonometric equation 4sin⁡2x−3=04\sin^2 x - 3 = 0?

  1. A
    {60∘}\{60^\circ\}
  2. {60∘,120∘}\{60^\circ, 120^\circ\}Cevap
  3. C
    {30∘,150∘}\{30^\circ, 150^\circ\}
  4. D
    {60∘,240∘}\{60^\circ, 240^\circ\}

Cevap

{60∘,120∘}\{60^\circ, 120^\circ\}
Solving the equation 4sin⁡2x−3=04\sin^2 x - 3 = 0 gives sin⁡2x=34\sin^2 x = \frac{3}{4}, which simplifies to sin⁡x=32\sin x = \frac{\sqrt{3}}{2} within the interval 0∘≤x≤180∘0^\circ \le x \le 180^\circ. The principal angle is 60∘60^\circ. Since sine is positive in both the first and second quadrants, the second valid angle in the domain is 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ, giving the solution set {60∘,120∘}\{60^\circ, 120^\circ\}.

Adım Adım Çözüm

1
Isolate the trigonometric term
sin⁡2x=34\sin^2 x = \frac{3}{4}
Rearrange 4sin⁡2x−3=04\sin^2 x - 3 = 0 by adding 3 to both sides and dividing by 4.
2
Take the square root of both sides
sin⁡x=±32\sin x = \pm \frac{\sqrt{3}}{2}
Taking the square root yields both positive and negative ratios.
3
Apply the domain restriction 0∘≤x≤180∘0^\circ \le x \le 180^\circ
sin⁡x=32\sin x = \frac{\sqrt{3}}{2}
The sine function is non-negative in the first and second quadrants (0∘≤x≤180∘0^\circ \le x \le 180^\circ), so the negative root has no solutions in this interval.
4
Determine the angles for xx
x = 60^\circ \text{ and } x = 180^\circ - 60^\circ = 120^\circ
The reference angle is 60∘60^\circ because sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}. In Quadrant II, the corresponding angle is 180∘−60∘=120∘180^\circ - 60^\circ = 120^\circ.

Anahtar Kavram

Solving Quadratic Trigonometric Equations
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Konular

Trigonometric Graphs and Simple Equations

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