7.8+OFSA+Solutions


 * OFSA SOLUTIONS for Solving Equations Using Trigonometric IDs**

This page contains peer generated solutions and error explanations to OFSA questions. As you read or view the solutions, be critical: check for accuracy, but also for more efficient solution strategies. If you have a better method or different idea/answer, post a discussion and monitor the responses.


 * Quick Directions**
 * Post answers, solutions and error explanations to each OFSA question below.
 * For each "distractor" or incorrect answer choice, explain the error that would lead to that incorrect answer choice.
 * You may either do the above in typed format or using a pencast.
 * Separate each question with a section bar.
 * After each solution, provide a hyperlink back to the corresponding OFSA page.
 * Follow example below.
 * Click here to refer to solution format in 7.7

math \[Solve\;for\;3tan^2x-sec^2x-5=0\;over\;(-\infty ,\infty )\] math
 * Question 1**

math 3tan^2x-(tan^2x+1)-5=0\\3tan^2x-tan^2x-1-5=0\\2tan^2x-6=0\\2(tan^2x-3)=0\\tan^2x=3\\x=\pm \sqrt{3}\\x=\frac{\pi}{3}+\pi n, \forall n\epsilon \mathbb{Z}\;or\;x=\frac{2\pi }{3}+\pi n, \forall n\epsilon \mathbb{Z}\] math
 * __Solution 1__**

math A.\;\;\;\;\[x=\frac{\pi }{6}+\pi n, \forall n\epsilon \mathbb{Z}\;or\;x=\frac{7\pi}{6}+\pi n, \forall n\epsilon \mathbb{Z}\\ {\text{\[Error. \;Student\;used\;incorrect\;unit\;circle\;values\;for\;tan\pm \sqrt{3}.\]} math
 * __Error Explanation 1__**

math B.\;\;\;\;x=\frac{\pi }{3}+\pi n, \forall n\epsilon \mathbb{Z}\\ {\text{Error. Student only accounted for the unit circle values for positive} \;tan\sqrt{3} \;{\text{and not negative} \;tan\sqrt{3}. math

math C.\;\;\;\;x=\frac{\pi }{3}+\pi n, \forall n\epsilon \mathbb{Z}\;or\;x=\frac{2\pi }{3}+\pi n, \forall n\epsilon \mathbb{Z}\\{\text{Correct!} math

math D.\;\;\;\;x=\frac{7\pi}{6}+\pi n, \forall n\epsilon \mathbb{Z}\\ {\text{Error. Student used incorrect unit circle values for} \;tan\pm \sqrt{3}\;{\text{and only accounted for the unit circle values for positive}\;tan\sqrt{3}\;{\text{and not negative}\; tan\sqrt{3}. math

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math Solve\;for\;tan^2x+2tanx=3\;over[0,2\pi ] math math \[tan^2x+2tanx-3=0\\(tanx-1)(tanx+3)=0\\tanx=1\;or\;tanx=-3\\x=Tan^{-1}(-1)\;or\;x=Tan^{-1}(-3)\\x=\frac{\pi}{4},\frac{5\pi}{4},\pi+Tan^{-1}(-3),2\pi+Tan^{-1}(-3)\] math math \[A.\;\;\;\;x=\frac{\pi}{4},\frac{5\pi}{4},\pi+Tan^{-1}(-3),2\pi+Tan^{-1}(-3)\]\\Correct! math
 * Question 2**
 * __Solution 2__**
 * __Error Explanation 2__**

math \[B.\;\;\;\;x=\frac{\pi}{4},\frac{5\pi}{4},\pi-Tan^{-1}(-3),2\pi-Tan^{-1}(-3)\]\\Error. Student substracted Tan^{-1}(-3) from\; \pi \;and \;2\pi \; rather \;than\; adding. \;By\; subtracting\; a\; negative\; angle, \;you \;would\; actually\; be\; adding\;Tan^{-1}(-3),\; which\; is\; incorrect\; because\;\pi-Tan^{-1}(-3)\; and\;2\pi-Tan^{-1}(-3)\; would\;result \;in \;angles\; located\; in\; quadrants\; III\; and\; I\; rather\; than\; quadrants\; II\; and\; IV. \;Also,\;2\pi-Tan^{-1}(-3)\; would\; not\; be\; within\; the\; restricted\; domain. math math \[C.\;\;\;\;x=\frac{\pi}{4},\pi+Tan^{-1}(-3),2\pi+Tan^{-1}(-3) math Error. Student only included one tangent value, but tangent values correspond to two values on the unit circle.

math \[D.\;\;\;\;x=\frac{\pi}{4},\frac{5\pi}{4},Tan^{-1}(-3)+\pi,Tan^{-1}(-3)+2\pi\\{\text{Error. Adding}\;\pi\;{\text{to}\;Tan^{-1}(-3)\;{\text{results in an angle that is not equivalent to}\;\pi+Tan^{-1}.\;{\text{The same idea applies for}\;Tan^{-1}(-3)+2\pi\;{\text{versus}\;2\pi+Tan^{-1}(-3).\] math

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math \[Solve\;for\;6sin(\frac{1}{2}x)=3\;over\;[0,2\pi)\] math math \[\[\[sin(\frac{1}{2}x)=\frac{1}{2}\\\\let\;A=\frac{1}{2}x\\\\sinA=\frac{1}{2}x\\\\\frac{1}{2}x\;\epsilon \;{\frac{\pi}{6}}+2\pi n, \forall n\epsilon \mathbb{Z}\;and\;\frac{5\pi}{6}+2\pi n, \forall n\epsilon \mathbb{Z}\\\\x\;\epsilon\; \frac{\pi}{3}+4\pi n, \forall n\epsilon \mathbb{Z}\;and\;\frac{5\pi}{3}+4\pi n, \forall n\epsilon \mathbb{Z}\\\\x\;\epsilon\;\frac{\pi}{3},\frac{5\pi}{3}\;over\;[0,2\pi)\]\] math math \[A.\;\;\;\;x\;\epsilon\;\frac{\pi}{6},\frac{5\pi}{6}\;over\;[0,2\pi)\]\\Error. Student forgot to multiply equation by 2 to isolate x. math
 * Question 3**
 * __Solution 3__**
 * __Error Explanation 3__**

math \[B.\;\;\;\;x\;\epsilon\;\frac{\pi}{3},\frac{5\pi}{3}\;over\;[0,2\pi)\]\\Correct! math

math C.\;\;\;\;x\;\epsilon\;\frac{\pi}{3},\frac{5\pi}{3},\frac{13\pi}{3},\frac{17\pi}{3}\;over\;[0,2\pi)\\{\text{Error.\;Last\;two\;solutions\;are\;not\;within\;the\;given\;interval.} math

math \[D.\;\;\;\;x\;\epsilon\;\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}\;over\;[0,2\pi)\]\\Error. Student forgot to multiply equation by 2 to isolate x and the last two solutions are not withing the given interval. math

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math {\text{Solve for X: }} 2\sin^2 x+ 3\cos x -3= 0 {\text{ over the interval } [0,2\Pi). \\ math
 * Question 4**


 * __Solution 4__**

math \begin{gathered} 2(1-\cos^2 x) + 3\cos x -3=0 \hfill \\ 2-2\cos^2x+3\cos x -3=0 \hfill \\ 2\cos^2x-3\cos x+1=0 \hfill \\ (2\cos x -1) (\cos x -1)=0 \hfill \\ \cos x = \frac {1}{2} {\text{and}} \cos x =1 \hfill \\ x= 0, \frac{\Pi}{3}, \frac{5\Pi}{3} \hfill \\ \end{gathered} math


 * __Error Explanation 4__**

math \begin{array}{*{20}{l}} &{ 0, 2\Pi, \frac {\Pi}{3}& \cos x =\frac{1}{2}. \\ &{ 0, 2\Pi, \frac {\Pi}{3}, \frac{5\Pi}{3}}& \\ &{ 0, \frac{\Pi}{3}, \frac{5\Pi}{3}}& \\ &{ 0, 2\Pi}&{{\text{Error. Student forgot about the}} \sin x = \frac{1}{2}. \\ \end{array} math

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math 17\sin ^2x-3\cos ^2x=2 math math 17\sin ^2x-3\cos ^2x=2
 * Question 5**
 * Solve for X:**
 * __Solution 5__**

17\sin ^2x-3\left ( 1-\sin ^2x \right )=2

17\sin ^2x-3+3\sin ^2x=2

20\sin ^2x=5

sin ^2x=\frac{1}{4}

sin x=\pm \sqrt{\frac{1}{}4}

\pm \sqrt{\frac{1}{}4}=\pm \frac{1}{2}

\arcsin \left ( \frac{1}{2} \right )=\left ( \frac{\pi }{6}, \frac{5\pi }{6} \right )

\arcsin \left ( \frac{-1}{2} \right )=\left ( \frac{7\pi }{6}, \frac{11\pi }{6} \right ) math

math A. \left ( \frac{\pi }{6},\frac{5\pi }{6} \right ) math Error: Student forgot to solve for the positive and negative answer after taking the square root of 1/4
 * __Error Explanation 5__**

math B. \left (\frac{\pi }{6} ,\frac{5\pi }{6}, \frac{7\pi }{6},\frac{11\pi }{6} \right ) math Correct Answer math C. \pm \frac{1}{2} math Error: the answer itself can't be 1/2 or -1/2 because that isn't solving for x...the student must make sure if its's solving for x, that x is alone on one side of the =, to have an actual answer for what x equals. D. No solution Error: This is a distraction, it throws off the student if the answer they got doesnt match or something else

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Solve: math (tan\theta)(sin\theta)-(cot\theta)(sin\theta)+(tan\theta)-(cot\theta)=0\;over\; [-\pi,\pi] math math (tan\theta-cot\theta)(sin\theta+1)=0
 * Question 6**
 * __Solution 6__**

tan\theta=cot\theta

tan\theta= \frac{1}{tan\theta}

tan^{2}\theta=1

tan\theta=+1, -1

{\text{ }}\boxed {\theta=(-\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4})} math math \text{AND}

sin\theta=-1

\theta= -\frac{\pi}{2} math math \text{However, if you plug this solution back into the original equation, the equation becomes incorrect.} math

math \begin{array}{*{20}{l}} &{\theta = -\frac{\pi}{2}}& \sin\theta =-1. \\ &{\theta = (-\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4})}}& \\ &{\theta = (-\frac{3\pi}{4}, -\frac{\pi}{2} -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4})}}&{{\text{Error. The original equation would not be correct if you plug in}} -\frac{\pi}{2} \\ &{\theta = (-\frac{3\pi}{4}, \frac{\pi}{4})}&{{\text{Error. Student only solved for}} \tan\theta = +1}. \\ \end{array} math Click here to return to OFSA.
 * __Error Explanation 6__**

math \text{Solve for} {\theta} : math math sin\theta cos2\theta -cos\theta sin2\theta +3csc\theta =2 {\text{ over the interval } [0,4\pi). \\ math
 * Question 7**

math sin(\theta -2\theta )+3csc\theta =2 math
 * __Solution 7__**

math sin(-\theta )+3csc\theta =2 math

math -sin\theta +\frac{3}{sin\theta }=2 math

math \frac{-sin^{2}\theta +3}{sin\theta }=2 math

math -sin^{2}\theta +3=2sin\theta math

math sin^{2}\theta +2sin\theta -3=0 math

math (sin\theta +3)(sin\theta -1)=0 math

math sin\theta =-3 math

math \theta =Sin^{-1}(-3)) math math \text{However, this solution is not on the unit circle, making this solution invalid.} math

math \text {AND} math

math sin\theta =1 math

math {\text{ }}\boxed {\theta= (\frac{\pi}{2}, \frac{5\pi}{2})} math

math \begin{array}{*{20}{l}} &{\theta = \frac{\pi}{2}}&{{\text{Error. Student only solved over the interval}}&[0,2\pi] \\ &{\theta = (Sin^{-1} (-3))}}& \\ &{\theta = (\frac{\pi}{2}, \frac{5\pi}{2}})}}&{{\text{Correct!} \\ &{\theta = \frac{5\pi}{2}}}&{{\text{Error. Student only solved over the interval}}&[2\pi, 4\pi)\\ \end{array} math Click here to return to OFSA.
 * __Error Explanation 7__**

math \sin \left ( 105 \right )
 * Question 8**

\sin \left ( 45+60 \right ) \sin \left ( 45 \right )\cos \left ( 60 \right )+\cos \left ( 45 \right )\sin \left ( 60 \right )

\left ( \frac{\sqrt{2}}{2}\right )\left ( \frac{1}{2} \right )+\left ( \frac{\sqrt{2}}{2} \right )\left ( \frac{\sqrt{3}}{2} \right )

\left ( \frac{\sqrt{2}}{4}\right )+\left ( \frac{\sqrt{6}}{4} \right ) math math A. \left ( \frac{\sqrt{2}}{4}\right )+\left ( \frac{\sqrt{6}}{4} \right ) math Correct Answer math B. \left ( \frac{\sqrt{2}}{4}\right )-\left ( \frac{\sqrt{6}}{4} \right ) math Error: Student mixed up when to add and subtract, but since it's sine, it remains the same math C. \left ( \frac{1}{2}\right )+\left ( \frac{\sqrt{3}}{4} \right ) math Error: student did: math \sin \left ( a \right )\cos \left ( a \right )+\cos \left ( b \right )\sin \left ( b \right ) math instead of math \sin \left ( a \right )\cos \left ( b \right )+\cos \left ( a \right )\sin \left ( b \right ) math
 * __Solution 8__**

D. Not enough information given in order to solve Error: Distraction answer! Student atomatically assumed since it wasn't on the unit circle, more information was needed, when they didnt open their eyes and realize it was two values on the unit circle added together Click here to return to OFSA.
 * __Error Explanation 8__**

How many solutions are there going to be for x? math 4\sin ^2\left ( x \right )+1=4\sin \left ( x \right ), over:\left [ -\pi ,2\pi \right ] math math 4\sin ^2\left ( x \right )+1=4\sin \left ( x \right )
 * Question 9**
 * __Solution 9__**

4\sin ^2\left ( x \right )-4\sin \left ( x \right )+1=0

\left ( 2\sin \left ( x \right )-1 \right ) \left ( 2\sin \left ( x \right )-1 \right )

\left ( 2\sin \left ( x \right )-1 \right )

x=\left ( \frac{\pi }{4}, \frac{3\pi }{4} \right ) math therefore, 2 __**Error Explanation 9**__ A. 2 Correct answer

B. 4 Error: Student saw there was a restriction, but thought it was going to be a "normal" restriction of being over the span of 2π

C. 6 Error: Student thought the answer was positive and negative, there for having this many solutions

D. Infinite Number of Solutions Error: Distraction answer: student dident pay attention to the restrictions, therefore forgetting them as a whole Click here to return to OFSA.

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 * Question 10**
 * __Solution 10__**
 * __Error Explanation 10__**

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 * Question 11**
 * __Solution 11__**
 * __Error Explanation 11__**

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 * Question 12**
 * __Solution 12__**
 * __Error Explanation 12__**

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 * Question 13**
 * __Solution 13__**
 * __Error Explanation 13__**

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 * Question 14**
 * __Solution 14__**
 * __Error Explanation 14__**

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 * Question 15**
 * __Solution 15__**
 * __Error Explanation 15__**

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 * Question 16**
 * __Solution 16__**
 * __Error Explanation 16__**

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 * Question 17**
 * __Solution 17__**
 * __Error Explanation 17__**

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 * Question 18**
 * __Solution 18__**
 * __Error Explanation 18__**

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 * Question 19**
 * __Solution 19__**
 * __Error Explanation 19__**

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 * Question 20**
 * __Solution 20__**
 * __Error Explanation 20__**

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 * Question 21**
 * __Solution 21__**
 * __Error Explanation 21__**

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 * Question 22**
 * __Solution 22__**
 * __Error Explanation 22__**

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 * Question 23**
 * __Solution 23__**
 * __Error Explanation 23__**

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 * Question 24**
 * __Solution 24__**
 * __Error Explanation 24__**