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Question Video: Finding the Solutions to a Trigonometric Equation over a Specified Range by Taking Square Roots Mathematics • 10th Grade

Find the set of values satisfying 4 sinΒ² πœƒ βˆ’ 1 = 0 given that 0 < πœƒ < 180Β°.

05:43

Video Transcript

Find the set of values satisfying four times the sin of πœƒ squared minus one equals zero given that πœƒ is greater than zero degrees and less than 180 degrees.

Let’s begin by solving for sin of πœƒ. We add one to each side of the equation. Now, four times the sin of πœƒ squared equals one. Then, we divide each side by four. So the sin of πœƒ squared equals one-fourth. It follows that sin of πœƒ equals the positive and negative square roots of one-fourth. To simplify the resulting square root of a quotient, we square root both the numerator and the denominator. Thus, sin of πœƒ equals positive and negative one-half. Because of the periodic nature of the sine function, this equation would have infinitely many solutions if we did not have restrictions on πœƒ. We are told that πœƒ is greater than zero degrees and less than 180 degrees.

We now recall the familiar graph of sine of an angle plotted against that angle. We see that sin of πœƒ equals one-half two times in the open interval from zero to 180 degrees. We have marked these places with blue crosses. However, since the sin of πœƒ never equals negative one-half between zero and 180 degrees, we have no solutions from the equation sin of πœƒ equals negative one-half.

Now we must find the two values of πœƒ for which the sin of πœƒ equals one-half. The exact solutions are not obvious from the graph. However, we see that one solution is less than 90 degrees and the other solution is closer to 180 degrees. However, by applying the inverse sine function to our equation, we obtain that the equation πœƒ equals sin inverse of one-half. Then, we can use a calculator in degree mode to find the first answer of 30 degrees. Unfortunately, by applying inverse sine, we are only able to find one of the two solutions between zero and 180 degrees. Inverse sine is only defined between negative 90 and 90 degrees. That is why it only returns one value of πœƒ.

To find the second value of πœƒ, we will need to recall the relationship between the sides of a 30-60-90 special right triangle. Let’s clear some space to show our reasoning. For this demonstration, we sketch a 30-60-90 reference triangle on the π‘₯𝑦-coordinate plane, where the hypotenuse is the terminal side of a 30-degree angle in standard position. According to coordinate definitions of trigonometric functions, if we let π‘Ÿ, the hypotenuse, equal one, then sin equals π‘₯ and cos equals 𝑦. Within a 30-60-90 triangle, the side opposite the 30-degree angle is always half the length of the hypotenuse. And the side adjacent to the 30-degree angle is always the square root of three times the length of the opposite side. Therefore, 𝑦 equals one-half and π‘₯ equals the square root of three over two. This reference triangle we have drawn confirms that sin of 30 degrees equals one-half.

Now we will look in the second quadrant for our second solution, where πœƒ is between 90 and 180 degrees. We sketch another 30-60-90 reference triangle in the second quadrant along the negative π‘₯-axis, with the same measures as our first right triangle. Since π‘₯-values in the second quadrant are always negative, we now have π‘₯ equal to negative square root of three over two. But more importantly, the value of 𝑦 remains the same. Therefore, sine still equals one-half.

The reference angle in the second quadrant is still 30 degrees. But we need to identify the measure of this angle in standard position. We know that an angle terminating on the negative side of the π‘₯-axis equals 180 degrees. Our angle is a 30-degree clockwise turn from the negative π‘₯-axis. This reminds us of one of the cofunction identities of sine, which tells us the sin of πœƒ equals sin of 180 degrees minus πœƒ. The hypotenuse of our first triangle is the terminal side of angle πœƒ, and the hypotenuse of our second triangle is the terminal side of the 180 degrees minus πœƒ angle.

We found πœƒ in the first quadrant to equal 30 degrees. So, according to the cofunction identify we just recalled, sin of 30 degrees equals sin of 180 degrees minus 30 degrees, which is 150 degrees. We have demonstrated that both sin of 30 degrees and sin of 150 degrees equal one-half. In conclusion, the set containing 30 degrees and 150 degrees satisfies the equation four times the sin of πœƒ squared minus one equals zero, given that πœƒ is greater than zero degrees and less than 180 degrees.

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