Question textSolve the equation [math: 4sin⁡(x)+2=0]{4\,\sin \left( x \right)+2=0} for [math: x∈[0,2π]]\displaystyle x\in \Big[0, 2\pi \Big], giving your answer in radians. Note. Type pi to enter [math: π]\pi and write your answers [math: a] and [math: b] in the form [math: {a,b}]\{a,b\}. [input] Check Question 1

Question

BlackTom AI · Accepted Answer

We start by restating the problem in our own words to focus on what is being asked: solve the equation 4 sin(x) + 2 = 0 for x in [0, 2π], and express the solutions in radians (using pi notation for multiples of π).

First, isolate the sine term: 4 sin(x) + 2 = 0 ⇒ 4 sin(x) = -2 ⇒ sin(x) = -1/2. The task now is to find all angles x in the interval [0, 2π] where the sine value is -1/2.

Next, recall the unit circle values: sin(x) = -1/2 occurs in the third and fourth quadrants at the standard reference angle where sin equals 1/2, which is π/6. The corresponding angles in [0, 2π] are found by adding π to the reference angle for the third quadrant and 2π − π/6 for the fourth quadrant, or more generally by using the standard solutions for sin(x) = -1/2:
- x = π + π/6 = 7π/6
- x = 2π − π/6 = 11π/6

Now verify that these values lie within the interval [0, 2π]: both 7π/6 and 11π/6 are between 0 and 2π, so they are valid solutions.

To recap the reasoning steps in a compact form: solve sin(x) = -1/2, identify the quadrants where sine is negative (III and IV), determine the reference angle π/6, and map to the interval [0, 2π] to obtain the specific angles 7π/6 and 11π/6.

If you were entering these answers in a system that uses pi for π, you would express them as x = 7π/6 and x = 11π/6, which correspond to the two angles in the specified range where the sine value matches -1/2.

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Question textSolve the equation [math: 3tan⁡(x)−3=0]{3\,\tan \left( x \right)-\sqrt{3}=0} over [math: [0,2π]]\displaystyle \Big[0, 2\pi \Big] in radians. Note. Type pi to enter [math: π]\pi and use { , } to enter a set.[input] Check Question 4

Question textSolve the equation [math: tan⁡(x)−3=0]{\tan \left( x \right)-\sqrt{3}=0} over [math: [0,2π]]\displaystyle \Big[0, 2\pi \Big] in radians. Note. Type pi to enter [math: π]\pi and use { , } to enter a set.[input] Check Question 4

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