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Chapter 19: Problem 1879

The number of values of \(\theta\) in the interval \([0,4 \pi]\) satisfying the equation \(2 \sin ^{2} \theta-\cos 2 \theta=0\) (a) 4 (b) 8 (c) 2 (d) 6

Short Answer

Expert verified
There are 6 values of \(\theta\) in the interval \([0, 4\pi]\) that satisfy the equation \(2\sin^2{\theta}-\cos{2\theta}=0\). These values are \(\frac{\pi}{6}, \pi-\frac{\pi}{6}, \frac{7\pi}{6}, 2\pi-\frac{\pi}{6}, \pi+\frac{\pi}{6},\) and \(3\pi+\frac{\pi}{6}\). Therefore, the correct answer is (d) 6.

Step by step solution

01

Simplify the equation

Recall the double-angle formula for cosine, \(\cos{2\theta}=1-2\sin^2{\theta}\). We can rewrite the given equation as follows: \[2\sin^2{\theta}-(1-2\sin^2{\theta})=0\]
02

Solve for \(\sin{\theta}\)

Continuing from Step 1, now solve for \(\sin{\theta}\): \[2\sin^2{\theta}+(2\sin^2{\theta}-1)=0\] \[4\sin^2{\theta}-1=0\] \[\sin^2{\theta}=\frac{1}{4}\] \[\sin{\theta}=\pm\frac{1}{2}\]
03

Find solutions for \(\theta\)

Now we have sin of \(\theta\), we need to find the angle \(\theta\) in radians: \[\theta_1=\arcsin{\frac{1}{2}}\] \[\theta_2=\arcsin{-\frac{1}{2}}\] For \(\theta_1\), we have: \[\theta_1=\arcsin{\frac{1}{2}}=\frac{\pi}{6}\] We know that \(\sin{\theta}\) is positive in the 1st and 2nd quadrants. Therefore, \[\theta_1=\frac{\pi}{6}+2\pi k\] \[\theta_1=\pi-\frac{\pi}{6}+2\pi k\] Where \(k\) is an integer. Similarly, for \(\theta_2\), we have: \[\theta_2=\arcsin{-\frac{1}{2}}=-\frac{\pi}{6}\] We know that \(\sin{\theta}\) is negative in the 3rd and 4th quadrants. Therefore, \[\theta_2=\pi+\frac{\pi}{6}+2\pi k\] \[\theta_2=2\pi-\frac{\pi}{6}+2\pi k\] Where \(k\) is an integer.
04

Count the solutions in the interval \([0, 4\pi]\)

Finally, we will count the number of solutions in the given interval for each one of the general solutions \(\theta_1\) and \(\theta_2\): For \(\theta_1\), we have: \[\theta_1=\frac{\pi}{6},\pi-\frac{\pi}{6},\frac{7\pi}{6},2\pi-\frac{\pi}{6}\] For \(\theta_2\), we have: \[\theta_2=\pi+\frac{\pi}{6},2\pi-\frac{\pi}{6},3\pi+\frac{\pi}{6},4\pi-\frac{\pi}{6}\] Our interval is \([0,4\pi]\). The last solution, \(4\pi-\frac{\pi}{6}\), is greater than \(4\pi\). So we have only 7 solutions that satisfy the equation in the interval \([0,4\pi]\): \[\theta={\frac{\pi}{6},\pi-\frac{\pi}{6},\frac{7\pi}{6},2\pi-\frac{\pi}{6},\pi+\frac{\pi}{6},2\pi-\frac{\pi}{6},3\pi+\frac{\pi}{6}}\] The correct answer is (d) 6.

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