Question textSolve the equation [math: 2sin⁡(x)+1=0]{\sqrt{2}\,\sin \left( x \right)+1=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

The problem asks to solve the equation sqrt(2) sin(x) + 1 = 0 for x in [0, 2π], with the answer to be entered in radians in the form {a,b}.
First, isolate sin(x): sqrt(2) sin(x) + 1 = 0 ⇒ sqrt(2) sin(x) = -1 ⇒ sin(x) = -1 / sqrt(2) = -√2/2.
Next, think about where sine takes the value -√2/2 on the interval [0, 2π]. On the unit circle, sin(x) = -√2/2 occurs at angles in the third and fourth quadrants where the reference angle is π/4 (since sin(π/4) = √2/2).
Therefore, the general solutions in [0, 2π) are x = π + π/4 = 5π/4 and x = 2π - π/4 = 7π/4.
Ordering them from smallest to largest within the interval, we have x = 5π/4 and x = 7π/4. In the requested form {a,b}, this corresponds to {5π/4, 7π/4}.
Note: If one mistakenly solves a different coefficient, such as 2 sin(x) + 1 = 0, you would obtain sin(x) = -1/2, which yields x = 7π/6 and x = 11π/6, not the values above. The given equation, however, uses sqrt(2) sin(x) + 1 = 0, so the correct solutions are the ones identified here.
Be mindful of input conventions: the form {a,b} should reflect the two values in ascending order unless the system specifies a different ordering rule.

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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

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