It’s your turn on trigonometric inequalities, 1

It's your turn on Trigonometric inequalities
11.1
Welcome again to “It’s Your Turn” step. You’re dealing today with trigonometric inequalities, which is quite a difficult topic. In exercise one, we’ve got a product of terms, and we want it to be strictly positive. So we will study separately the sign of the two terms. Let us begin with the study of the sign of sine of x. Well, sine of x vanishes whenever x is a multiple of pi.
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And we can indicate the zeros of sine x in the trigonometric circle, and corresponding to 0 and pi. We shall use the inner circle for the sign of sine of x. So we write here 0, and we write here 0. And these will be used for the sign of sine of x. So the sine of x is positive here, and negative elsewhere And concerning 2 cosine of x minus 1, this is equal to 0 whenever cosine of x equals to 1/2. And this means, well, x equals pi third, or minus pi divided by 3, plus multiple of pi. Again, we can indicate these values in the outer trigonometric circle.
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And here we have got pi thirds, and here we’ve got minus pi thirds.
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The cosine of x is strictly greater than 1/2 whenever we are on the right-hand side of this vertical line.
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And so the cosine will be positive in this region. So here, will be greater than 1/2 in this region. And we write the plus sign to the expression 2 cosine of x minus 1. And here we write the negative sign.
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So here we have a negative sign, negative sign, negative sign, and a negative sign. Now we use the product rule of the signs, and we write in the outer region the sign of sine of x times 2 cosine of x minus 1. Here, in this region, we get a plus sign.
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Here we get 0. Here also we get 0, because it is a zero of 2 cosine x minus 1. Here also we get 0– let me write it here– because it is a zero of 2 cosine of x minus 1. Here we get minus times plus. So a minus sign. Here we get the plus sign.
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Here we get 0, because it is a zero of sine of x. And here we get again a minus sign.
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So the regions where the product is positive is this region and this other region here. So the solution sine of x times 2 cosine of x minus 1 will be strictly greater than 0 if and only if x belongs– well, here we are in minus pi up to minus pi third, or from 0 to pi third plus any multiple of 2 pi.
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And this ends exercise one.

Do your best in trying to solve the following problems. In any case some of them are solved in the video and all of them are solved in the pdf file below.

Exercise 1.

Solve the inequality (sin xleft(2cos x-1right)>0. %, b) frac{4sin^2(x)-1}{2cos(x)}geq 0.)