Find all solutions to the equation.3tanθ−7=−6\sqrt3\tan\theta-7=-63tanθ−7=−6

θ = π 6 + π n \theta=\frac{\pi}{6}+\pi n θ = 6 π + πn

Find all solutions to the equation. 3 tan θ − 7 = − 6 \sqrt3\tan\theta-7=-6 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> tan θ − 7 = − 6

In this problem, we're asked to find all solutions to the equation, the square root of 3 times tangent of theta minus 7 is equal to negative 6. So let's go ahead and get started with step number 1 and isolate our trig function. So here, we wanna isolate the trig function, the tangent of theta, and the first step in doing that is adding 7 to both sides. So we'll cancel on that left side, leaving me with the square root of 3 times the tangent of theta is equal to negative 6 +7, which will give me a positive one. Now from here to fully isolate my trig function, I can divide both sides by the square root of 3. Now that'll, of course, cancel on that left side, leaving me with the tangent of theta is equal to 1 over the square root of 3. Now this value might not immediately look familiar because we can go ahead and rationalize this denominator, which we can do by multiplying both the numerator and denominator by the square root of 3. Now in my numerator, I'm left with the square root of 3, and in my denominator, I am left with just 3. So here, my final equation with my trig function isolated is going to be the tangent of theta is equal to the square root of 3 over 3, which is a much more familiar value. Now we've completed step 1 here. We can move on to step number 2 and find all of our solutions on the unit circle. So coming down to our unit circle here, we wanna find all of our angles theta for which the tangent is equal to the square root of 3 over 3. Now because this is a positive value, I know that my tangent values are are positive in quadrant 1 and also in quadrant 3. So here in quadrant 1, I see that for my angle pi over 6, my tangent is equal to root 3 over 3. So that is one of the solutions on my unit circle. And over here in quadrant 3, for my angle 7pi over 6, my tangent will also be equal to the square root of 3 over 3. So for step 2, my solutions that are on my unit circle are theta equals pi over 6 and theta equals 7 pi over 6. So here we have completed step number 2. Moving on to step number 3, here we're told that if our domain is not restricted, we should add 2 pi into each solution. Now here in this problem, we were told to find all solutions to this equation, and we're not given a restriction. So we need to go ahead and add that 2 pi n to each of these solutions. So here I have my solutions, theta equals pi over 6 plus 2 pi n, and theta equals 7 pi over 6 plus 2 pi n. So we have completed step number 3. Now step number 4 is to isolate theta, but theta is already isolated here. We already have theta by itself, so these are actually our final solutions, and step number 4 is done. But remember, when working with the tangent of an angle, you'll see this thing happen where our angles on our unit circle are exactly half a rotation apart. They are pi radians apart. So we can actually further simplify these solutions. Instead of representing them in 2 different expressions, we can actually go ahead and kind of combine them. Because if I have my answer as theta equals pi over 6+in, n. Looking at that angle pi over 6 on my unit circle here, starting at pi over 6, if I go pi radians, I reach the second solution on my unit circle, 7 pi over 6. Then if I come around another pi radians, I'm back at that pi over 6, and that will continue on and on and on, giving me all of my solutions. Now this only works for the tangent. So whenever you're working with the tangent, you can always simplify your expression or your solutions to be just adding pi n instead of adding 2 pi n to both of them. So here we have our final fully simplified solution as theta equals pi over 6+pi n radians. Thanks for watching, and I'll see you on the next one.

Find all solutions to the equation. 3 tan ⁡ θ − 7 = − 6 \sqrt3\tan\theta-7=-6 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> tan θ − 7 = − 6

θ = π 6 + π n \theta=\frac{\pi}{6}+\pi n θ = 6 π + πn

θ = π 6 + 2 π n , 5 π 6 + 2 π n \theta=\frac{\pi}{6}+2\pi n,\frac{5\pi}{6}+2\pi n θ = 6 π + 2 πn , 6 5 π + 2 πn

θ = 5 π 6 + 2 π n , 11 π 6 + 2 π n \theta=\frac{5\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n θ = 6 5 π + 2 πn , 6 11 π + 2 πn

θ = π 6 + 2 π n , 7 π 6 + 2 π n \theta=\frac{\pi}{6}+2\pi n,\frac{7\pi}{6}+2\pi n θ = 6 π + 2 πn , 6 7 π + 2 πn

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Linear Trigonometric Equations

Trigonometry

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Linear Trigonometric Equations

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Multiple Choice

Find all solutions to the equation.
$\sqrt3\tan\theta-7=-6$

$\theta=\frac{\pi}{6}+2\pi n,\frac{5\pi}{6}+2\pi n$

$\theta=\frac{5\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n$

$\theta=\frac{\pi}{6}+2\pi n,\frac{7\pi}{6}+2\pi n$

$\theta=\frac{\pi}{6}+\pi n$

Verified step by step guidance

Start by isolating the tangent term in the equation: 3\tan\theta - 7 = -6. Add 7 to both sides to get 3\tan\theta = 1.

Divide both sides by 3 to solve for \tan\theta: \tan\theta = \frac{1}{3}.

Recall that the general solution for \tan\theta = a is \theta = \arctan(a) + \pi n, where n is an integer, because the tangent function has a period of \pi.

Calculate \arctan(\frac{1}{3}) to find the principal value of \theta. This gives \theta = \frac{\pi}{6} as one solution.

Since the period of the tangent function is \pi, the general solution is \theta = \frac{\pi}{6} + \pi n, where n is an integer.

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