Mathematics, 02.08.2019 02:40 archersmithdrag

What values for theta (0 less than or equal to theta less than equal to 2pi) satisfy the equation? 2 sin theta cos theta + sqrt 3 cos theta = 0 the answer is pi/2, 4pi/3, 3pi/2, 5pi/3 i just need to know how to get it

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Answer from: tacie57

$2\sin\theta \cos \theta + \sqrt{3}\cos\theta = 0$ for $0\leq\theta\leq2\pi$ cosine theta can be factored out. all it is doing is multiplying with multiple terms after all. $\cos \theta\left(2\sin\theta + \sqrt{3}\right) = 0$ now we apply the zero product property.the zero product property allows us to say that either $\cos \theta$ or $\left(2\sin\theta + \sqrt{3}\right)$ can be zero for the equation to be true.it's just like how in quadratics, when seeing $(x-1)(x-2)=0$ , the solutions are x=1 and x=2 because either one of those factors can be zero to satisfy the equation.if we focus on $\cos \theta$ , set it equal to zero and solve. $\cos \theta = 0$ you need to use either unit circle knowledge or recall what the graph of cosine looks like.since cosine is defined to be, for a standard position angle $\theta$ that has a point (x,y) on the terminal arm with r being the distance from the origin, we have $\cos\theta = x/r$ the places where we can get x = 0 are when the terminal arm of the angle is pointing directly up (pi/2) and when the terminal arm is pointing directly down (3pi/2). we cannot go past 2π due to the given restrictions on this question.so the solutions for $\cos \theta = 0$ are $\theta = \pi/2, 3\pi/2$ now, we have one other factor to consider: $\left(2\sin\theta + \sqrt{3}\right)$ set that equal to zero and solve. $\begin{aligned} 2\sin\theta + \sqrt{3} & = 0\\ 2\sin\theta & = -\sqrt{3} \\ \sin\theta & = -\frac{\sqrt{3}}{2} \end{aligned}$ from unit circle or the 30-60-90 triangle, we know that if we have a reference angle of $\pi/3$ , we will get a sine ratio of $\sqrt{3}/2$ .sine is negative. sine is only negative in quadrants 3 and 4.from the unit circle,the angle in quadrant 3 with a sine ratio of $-\sqrt{3}/2$ is $4\pi/3$ .the angle in quadrant 4 with a sine ratio of $-\sqrt{3}/2$ is $5\pi/3$ .there are no other solutions because we cannot go past 2π.therefore, the solutions to $\sin\theta = -\frac{\sqrt{3}}{2}$ are $\theta = 4\pi/3,\ 5\pi/3$ bringing solutions from both factors, we have $\theta = \dfrac{\pi}{2}, \dfrac{4\pi}{3}, \dfrac{3\pi}{2}, \dfrac{5\pi}{3}$

Answer from: Quest

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

Answer from: Quest

1.5 pounds because you put 9over 6 and then divide by 6

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