### Video Transcript

Find all points with π₯-coordinates in the interval from zero to π where the curve π¦ equals sin two π₯ has a tangent that is parallel to the line π¦ equals negative π₯ minus 18.

We recall first that two lines are parallel if they have the same slope. We can determine the slope of the line π¦ equals negative π₯ minus 18 as the equation of this line has been given in the slopeβintercept form. Remember the slopeβintercept form of the equation of a straight line is π¦ equals ππ₯ plus π, where π represents the slope of the line and π represents its π¦-intercept. Comparing the equation of this line with the slopeβintercept form, we see that the coefficient of π₯, and hence the slope of the line, is negative one. We now know that weβre looking for the points on the curve π¦ equals sin two π₯ where the tangent to the curve has a slope equal to negative one.

Remember that the slope of the tangent to a curve is the same as the slope of the curve itself at that point. We can find the slope function of a curve by using differentiation. The slope function is given by dπ¦ by dπ₯. We need to recall how we differentiate sin two π₯, which is a trigonometric function. We should recall that the derivative with respect to π₯ of sin of ππ₯, where π is a real constant and π₯ is measured in radians, is equal to π multiplied by cos of ππ₯. So applying this rule, we have that dπ¦ by dπ₯ is equal to two multiplied by cos of two π₯.

Now remember weβre looking for the points where the curve has a tangent that is parallel to the line π¦ equals negative π₯ minus 18. So weβre looking for the points where this slope function is equal to negative one. We can therefore set this expression equal to negative one and solve the equation. Dividing both sides by two gives cos of two π₯ is equal to negative one-half. To find the value of two π₯, we need to apply the inverse cosine function, giving two π₯ equals cos inverse of negative one-half.

We can evaluate this in our calculators, making sure that theyβre in radian mode. And it gives two π₯ is equal to two π by three. We can then divide both sides of the equation by two, giving π₯ equals π by three. So we found the π₯-coordinate of one point, where the tangent is parallel to the given line. But are there any others in the interval from zero to π? Now if π₯ belongs in the interval from zero to π, which is closed at the lower end and open at the upper end, then two π₯ will belong in the interval from zero to two π.

Recalling the graph of the cosine function between zero and two π, then we see that there are two values of π in this interval for which cos of π is equal to negative one-half. Weβre using π here as a placeholder for two π₯. Now weβve already worked out one of these values of two π₯ was two π by three. And to work out the other value, we can recall the symmetry of the cosine function. For values of π between zero and two π, cos of π is equal to cos of two π minus π. So to find the second solution, we can subtract the value weβve already found of two π by three from two π.

We can write two π as six π over three, so we have two π₯ equals six π over three minus two π over three, which is four π over three. Remember this is the value of two π₯. So to find the value of π₯, we divide both sides by two, giving π₯ equals two π by three. Both of these π₯-values are in the interval from zero to π. We now need to work out the values of π¦ that correspond to each of these π₯-values.

We work these out by substituting our π₯-values back into the equation of the function. When π₯ is equal to π by three, π¦ is equal to sin of two π by three, which is equal to root three over two. When π₯ is equal to two π by three, π¦ is equal to sin of two times two π by three or sin of four π by three, which is equal to negative root three over two. And so we found the coordinates of the two points in the given interval where the curve π¦ equals sin two π₯ has a tangent that is parallel to the line π¦ equals negative π₯ minus 18. They are π by three, root three over two and two π by three, negative root three over two.