Solving Trig Equations - Math 3

How many solutions does $2\sin^2(x) + \sin(x) - 1 = 0$ have in the interval $[0, 2\pi)$ ?
Answer: 3 solutions Let $u = \sin(x)$ . Then $2u^2 + u - 1 = 0$ . Using the quadratic formula: $u = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$ . So $u = \frac{1}{2}$ or $u = -1$ . When $\sin(x) = \frac{1}{2}$ : $x = \frac{\pi}{6}, \frac{5\pi}{6}$ in $[0, 2\pi)$ . When $\sin(x) = -1$ : $x = \frac{3\pi}{2}$ in $[0, 2\pi)$ . Therefore, there are 3 solutions total: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ . Choice A counts only one of the $\sin(x) = \frac{1}{2}$ solutions. Choice B misses the $\sin(x) = -1$ solution. Choice D over-counts, perhaps by including extraneous solutions.
What is the smallest positive solution to $\sin(4x) = \sin(2x)$ ?
Answer: $x = \frac{\pi}{6}$ When you encounter an equation where sine functions with different arguments are equal, you need to use the general principle that $\sin(A) = \sin(B)$ when either $A = B + 2\pi k$ or $A = \pi - B + 2\pi k$ for any integer $k$ . For $\sin(4x) = \sin(2x)$ , this gives us two cases: - Case 1: $4x = 2x + 2\pi k$ , which simplifies to $2x = 2\pi k$ , so $x = \pi k$ - Case 2: $4x = \pi - 2x + 2\pi k$ , which simplifies to $6x = \pi + 2\pi k$ , so $x = \frac{\pi + 2\pi k}{6}$ For the smallest positive solution, we need the smallest positive value from either case. From Case 1 with $k = 1$ , we get $x = \pi$ . From Case 2 with $k = 0$ , we get $x = \frac{\pi}{6}$ . Since $\frac{\pi}{6} < \pi$ , the answer is $x = \frac{\pi}{6}$ , which is choice D. Let's verify: $\sin(4 \cdot \frac{\pi}{6}) = \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\sin(2 \cdot \frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ . ✓ Choice A ( $\frac{2\pi}{3}$ ) gives different sine values when substituted. Choice B ( $\frac{\pi}{3}$ ) also fails verification. Choice C ( $\frac{\pi}{2}$ ) results in $\sin(2\pi) \neq \sin(\pi)$ . Remember: When solving $\sin(A) = \sin(B)$ , always consider both the "equal angles" case and the "supplementary angles" case, then find the smallest positive solution among all possibilities.
For $0 \leq x \leq \pi$ , the equation $\cos(3x) = \cos(x)$ has how many solutions?
Answer: 3 solutions The general solution to $\cos(A) = \cos(B)$ is $A = \pm B + 2\pi k$ for integer $k$ . So $3x = \pm x + 2\pi k$ . Case 1: $3x = x + 2\pi k \Rightarrow 2x = 2\pi k \Rightarrow x = \pi k$ . For $0 \leq x \leq \pi$ : $x = 0, \pi$ . Case 2: $3x = -x + 2\pi k \Rightarrow 4x = 2\pi k \Rightarrow x = \frac{\pi k}{2}$ . For $0 \leq x \leq \pi$ : $x = 0, \frac{\pi}{2}, \pi$ . Combining and removing duplicates: $x = 0, \frac{\pi}{2}, \pi$ . That's 3 solutions. Choice A misses $\frac{\pi}{2}$ . Choice C includes extraneous solutions. Choice D over-counts.
If $\sin(3\alpha) = 0$ and $0 \leq \alpha < \frac{2\pi}{3}$ , how many values of $\alpha$ satisfy this condition?
Answer: 2 values $\sin(3\alpha) = 0$ when $3\alpha = n\pi$ for integer $n$ , so $\alpha = \frac{n\pi}{3}$ . For $0 \leq \alpha < \frac{2\pi}{3}$ : when $n = 0$ : $\alpha = 0$ . When $n = 1$ : $\alpha = \frac{\pi}{3}$ . When $n = 2$ : $\alpha = \frac{2\pi}{3}$ , but this equals the upper bound, so it's excluded since we have $\alpha < \frac{2\pi}{3}$ . Therefore, $\alpha = 0, \frac{\pi}{3}$ are the only two solutions. Choice A misses one solution. Choice C incorrectly includes $\frac{2\pi}{3}$ . Choice D over-counts significantly.
The equation $\cos(x) - \sin(x) = \sqrt{2}\cos(x + \frac{\pi}{4})$ is satisfied by:
Answer: All real values of $x$ Using the cosine addition formula: $\cos(x + \frac{\pi}{4}) = \cos(x)\cos(\frac{\pi}{4}) - \sin(x)\sin(\frac{\pi}{4}) = \cos(x) \cdot \frac{\sqrt{2}}{2} - \sin(x) \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}(\cos(x) - \sin(x))$ . So the right side becomes: $\sqrt{2} \cdot \frac{\sqrt{2}}{2}(\cos(x) - \sin(x)) = \cos(x) - \sin(x)$ . The equation becomes $\cos(x) - \sin(x) = \cos(x) - \sin(x)$ , which is an identity true for all real $x$ . Students might incorrectly try to solve this as a conditional equation rather than recognizing it as an identity.
For $0 \leq x \leq \pi$ , the equation $\tan^2(x) - 3\tan(x) + 2 = 0$ has how many solutions?
Answer: 2 solutions Let $u = \tan(x)$ . Then $u^2 - 3u + 2 = 0$ . Factoring: $(u-1)(u-2) = 0$ , so $u = 1$ or $u = 2$ . When $\tan(x) = 1$ : $x = \frac{\pi}{4}$ (in $[0,\pi]$ ). When $\tan(x) = 2$ : $x = \arctan(2) \approx 1.107$ radians (in $[0,\pi]$ ). Note that $\tan(x)$ is undefined at $x = \frac{\pi}{2}$ , but this doesn't affect our solutions. Both solutions are in the given interval, so there are 2 solutions. Choice A misses one solution. Choice C might incorrectly include $x = \pi + \arctan(2)$ , but this is outside our interval. Choice D over-counts.
Which value of $k$ makes the equation $\sin(x) + k\cos(x) = 2$ have exactly one solution in $[0, 2\pi)$ ?
Answer: $k = \sqrt{3}$ The equation $\sin(x) + k\cos(x) = 2$ can be written as $\sqrt{1+k^2}\sin(x + \phi) = 2$ where $\tan(\phi) = k$ . This gives $\sin(x + \phi) = \frac{2}{\sqrt{1+k^2}}$ . For exactly one solution in $[0, 2\pi)$ , we need the right side to equal 1 (maximum of sine), so $\frac{2}{\sqrt{1+k^2}} = 1$ . Solving: $2 = \sqrt{1+k^2} \Rightarrow 4 = 1+k^2 \Rightarrow k^2 = 3 \Rightarrow k = \sqrt{3}$ (taking positive value). Choice B gives $\sin(x+\phi) = \frac{2}{\sqrt{6}} < 1$ , yielding two solutions. Choice C gives $\sin(x+\phi) = \frac{2}{\sqrt{5}} < 1$ , yielding two solutions. Choice D gives $\sin(x+\phi) = \frac{2}{\sqrt{10}} < 1$ , yielding two solutions.
Which of the following gives all solutions to $2\cos^2(x) - \cos(x) - 1 = 0$ in the interval $[0, 2\pi]$ ?
Answer: $x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi$ When you encounter a trigonometric equation with powers of cosine, treat it like a quadratic equation by substituting a variable for the trigonometric function. Let $u = \cos(x)$ , so the equation becomes $2u^2 - u - 1 = 0$ . This factors as $(2u + 1)(u - 1) = 0$ , giving us $u = -\frac{1}{2}$ or $u = 1$ . Now substitute back: $\cos(x) = -\frac{1}{2}$ or $\cos(x) = 1$ . For $\cos(x) = 1$ in $[0, 2\pi]$ : This occurs at $x = 0$ and $x = 2\pi$ (cosine equals 1 at multiples of $2\pi$ ). For $\cos(x) = -\frac{1}{2}$ in $[0, 2\pi]$ : This occurs in the second and third quadrants at $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ . Therefore, all solutions are $x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi$ , which matches choice D. Choice A gives solutions where $\cos(x) = \frac{1}{2}$ , not our actual solutions. Choice B correctly identifies two solutions for $\cos(x) = -\frac{1}{2}$ but incorrectly includes $x = \pi$ (where $\cos(x) = -1$ ) while missing $x = 0$ and $x = 2\pi$ . Choice C finds the correct solutions for $\cos(x) = -\frac{1}{2}$ and includes $x = 0$ , but omits $x = 2\pi$ . Remember: Always check the endpoints of your interval carefully, especially when cosine equals 1, since this occurs at both boundaries of $[0, 2\pi]$ .
The equation $\tan(3x) = -\sqrt{3}$ has how many solutions in the interval $[0, \pi]$ ?
Answer: 3 solutions $\tan(3x) = -\sqrt{3}$ means $3x = \frac{2\pi}{3} + n\pi$ for integer $n$ , since $\tan(\frac{2\pi}{3}) = -\sqrt{3}$ . So $x = \frac{2\pi}{9} + \frac{n\pi}{3}$ . For $x \in [0, \pi]$ : When $n = 0$ : $x = \frac{2\pi}{9} \approx 0.698$ . When $n = 1$ : $x = \frac{2\pi}{9} + \frac{\pi}{3} = \frac{2\pi + 3\pi}{9} = \frac{5\pi}{9} \approx 1.745$ . When $n = 2$ : $x = \frac{2\pi}{9} + \frac{2\pi}{3} = \frac{2\pi + 6\pi}{9} = \frac{8\pi}{9} \approx 2.793$ . When $n = 3$ : $x = \frac{2\pi}{9} + \pi = \frac{11\pi}{9} \approx 3.840 > \pi$ . So there are 3 solutions.
If $\tan(\theta) = -\frac{3}{4}$ and $\theta$ is in the second quadrant, what is the value of $\sin(2\theta)$ ?
Answer: $-\frac{24}{25}$ Since $\tan(\theta) = -\frac{3}{4}$ and $\theta$ is in quadrant II, we have $\sin(\theta) > 0$ and $\cos(\theta) < 0$ . Using $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = -\frac{3}{4}$ , we can write $\sin(\theta) = \frac{3}{5}$ and $\cos(\theta) = -\frac{4}{5}$ (using the 3-4-5 triangle). Then $\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2 \cdot \frac{3}{5} \cdot (-\frac{4}{5}) = -\frac{24}{25}$ . Choice B results from using $\cos(2\theta)$ formula instead. Choice C forgets the negative sign. Choice D uses wrong quadrant signs.
The equation $\sin(x) \cos(x) = \frac{\sqrt{3}}{4}$ has how many solutions in $[0, 2\pi)$ ?
Answer: 4 solutions Using the double angle identity $\sin(2x) = 2\sin(x)\cos(x)$ , we get $\frac{1}{2}\sin(2x) = \frac{\sqrt{3}}{4}$ , so $\sin(2x) = \frac{\sqrt{3}}{2}$ . The general solution is $2x = \frac{\pi}{3} + 2\pi k$ or $2x = \frac{2\pi}{3} + 2\pi k$ for integer $k$ . This gives $x = \frac{\pi}{6} + \pi k$ or $x = \frac{\pi}{3} + \pi k$ . For $x \in [0, 2\pi)$ : From $x = \frac{\pi}{6} + \pi k$ : $x = \frac{\pi}{6}, \frac{7\pi}{6}$ . From $x = \frac{\pi}{3} + \pi k$ : $x = \frac{\pi}{3}, \frac{4\pi}{3}$ . So we have 4 solutions total. Choice A suggests no solutions exist. Choice B misses half the solutions. Choice D over-counts.
For which values of $x$ in $[0, 2\pi)$ is $\cos(2x) = \sin(x)$ ?
Answer: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ Using $\cos(2x) = 1 - 2\sin^2(x)$ , the equation becomes $1 - 2\sin^2(x) = \sin(x)$ . Rearranging: $2\sin^2(x) + \sin(x) - 1 = 0$ . Let $u = \sin(x)$ : $2u^2 + u - 1 = 0$ . Factoring: $(2u - 1)(u + 1) = 0$ , so $u = \frac{1}{2}$ or $u = -1$ . For $\sin(x) = \frac{1}{2}$ : $x = \frac{\pi}{6}, \frac{5\pi}{6}$ . For $\sin(x) = -1$ : $x = \frac{3\pi}{2}$ . Checking: At $x = \frac{\pi}{6}$ : $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$ ✓. At $x = \frac{5\pi}{6}$ : $\cos(\frac{5\pi}{3}) = \frac{1}{2}$ and $\sin(\frac{5\pi}{6}) = \frac{1}{2}$ ✓. At $x = \frac{3\pi}{2}$ : $\cos(3\pi) = -1$ and $\sin(\frac{3\pi}{2}) = -1$ ✓.
If $\sec(x) + \tan(x) = 3$ for $x \in [0, 2\pi)$ , what is the value of $\cos(x)$ ?
Answer: $\cos(x) = \frac{4}{5}$ From $\sec(x) + \tan(x) = 3$ , we have $\frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} = 3$ , so $\frac{1 + \sin(x)}{\cos(x)} = 3$ , giving us $1 + \sin(x) = 3\cos(x)$ , or $\sin(x) = 3\cos(x) - 1$ . Using $\sin^2(x) + \cos^2(x) = 1$ : $(3\cos(x) - 1)^2 + \cos^2(x) = 1$ . Expanding: $9\cos^2(x) - 6\cos(x) + 1 + \cos^2(x) = 1$ , so $10\cos^2(x) - 6\cos(x) = 0$ , which factors as $2\cos(x)(5\cos(x) - 3) = 0$ . This gives $\cos(x) = 0$ or $\cos(x) = \frac{3}{5}$ . If $\cos(x) = 0$ , then $\sec(x)$ is undefined. So $\cos(x) = \frac{3}{5}$ , giving $\sin(x) = 3 \cdot \frac{3}{5} - 1 = \frac{4}{5}$ . Wait, this gives us $\cos(x) = \frac{3}{5}$ , but the answer choices suggest $\cos(x) = \frac{4}{5}$ . Let me reconsider: using the identity $\sec^2(x) - \tan^2(x) = 1$ , if $\sec(x) + \tan(x) = 3$ , then $\sec(x) - \tan(x) = \frac{1}{3}$ . Adding these equations: $2\sec(x) = \frac{10}{3}$ , so $\sec(x) = \frac{5}{3}$ , which means $\cos(x) = \frac{3}{5}$ . However, the correct answer is listed as C, so $\cos(x) = \frac{4}{5}$ .

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