Find all solutions to the equation.2cosθ+4=5\sqrt2\cos\theta+4=52cosθ+4=5

θ = π 4 + 2 π n , 7 π 4 + 2 π n \theta=\frac{\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n θ = 4 π + 2 πn , 4 7 π + 2 πn

Find all solutions to the equation. 2 cos θ + 4 = 5 \sqrt2\cos\theta+4=5 2 <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"> cos θ + 4 = 5

this problem, we're asked to find all solutions to the equation square root of 2 times cosine theta plus 4 is equal to 5. So let's get started with step number 1 and go ahead and isolate our trig function using whatever operations are needed. Now here I wanna isolate the cosine of theta, and I can start by simply subtracting 4 on both sides. Now that will cancel here, leaving me with root 2 cosine theta is equal to positive 1. Now from here, I can divide both sides by square root of 2, leaving me with cosine of theta on that left side, having fully isolated that, and then one over root 2 over here. Now that one over root 2 might not look super familiar to you, but remember that whenever we have a radical in the denominator, we can go ahead and rationalize that denominator. So in order to do that, I can multiply both sides or multiply the numerator and denominator by the square root of 2 over 2 because this is really just multiplying by 1. So that will leave me with a root 2 on the top and then just 2 on the bottom. Now this value might look a lot more familiar to you. So here I'm left with the cosine of theta is equal to root 2 over 2, and step number 1 is done. Now moving on to step 2, we wanna go ahead and find all of our solutions that are on our unit circle. So coming to my unit circle here, I wanna find all of the values for which or all of the angles for which my cosine is root 2 over 2. Now for pi over 4, I know that that is true. The cosine is root 2 over 2. And then also coming around to my 4th quadrant where my cosine values are positive yet again, I know that for 7 pi over 4, my cosine will also be root 2 over 2. So I have my solutions that are on my unit circle, pi over 4 and 7pi over 4. So from here, we have completed step number 2, and we have found all of our solutions on that unit circle. Now for step number 3, if our domain is not restricted, we wanna add 2 pi into each solution. Now in this problem, we were asked to find all solutions, which means that our domain is not restricted. So we can go ahead and add 2PI n to each of these values. Now from here step number 3 is done. We want to make sure that theta is all by itself. It's isolated representing our final answer, which in this case it is. So these are final solutions: pi over 4+2pinnand7pivour+2pinn. Thanks for watching and let me know if you have any questions.

Find all solutions to the equation. 2 cos ⁡ θ + 4 = 5 \sqrt2\cos\theta+4=5 2 <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"> cos θ + 4 = 5

θ = π 4 + 2 π n , 7 π 4 + 2 π n \theta=\frac{\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n θ = 4 π + 2 πn , 4 7 π + 2 πn

θ = π 4 + 2 π n , 3 π 4 + 2 π n \theta=\frac{\pi}{4}+2\pi n,\frac{3\pi}{4}+2\pi n θ = 4 π + 2 πn , 4 3 π + 2 πn

θ = 3 π 4 + 2 π n , 5 π 4 + 2 π n \theta=\frac{3\pi}{4}+2\pi n,\frac{5\pi}{4}+2\pi n θ = 4 3 π + 2 πn , 4 5 π + 2 πn

θ = 5 π 4 + 2 π n , 7 π 4 + 2 π n \theta=\frac{5\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n θ = 4 5 π + 2 πn , 4 7 π + 2 πn

11. Inverse Trigonometric Functions and Basic Trig Equations

Linear Trigonometric Equations

Precalculus

11. Inverse Trigonometric Functions and Basic Trig Equations

Linear Trigonometric Equations

Struggling with Precalculus?

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Multiple Choice

Find all solutions to the equation.
$\sqrt2\cos\theta+4=5$

$\theta=\frac{\pi}{4}+2\pi n,\frac{3\pi}{4}+2\pi n$

$\theta=\frac{\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n$

$\theta=\frac{3\pi}{4}+2\pi n,\frac{5\pi}{4}+2\pi n$

$\theta=\frac{5\pi}{4}+2\pi n,\frac{7\pi}{4}+2\pi n$

Verified step by step guidance

Start by simplifying the given equation: $ 2\cos\theta + 4 = 5 $. Subtract 4 from both sides to isolate the cosine term: $ 2\cos\theta = 1 $.

Divide both sides by 2 to solve for $ \cos\theta $: $ \cos\theta = \frac{1}{2} $.

Recall that the cosine function equals $ \frac{1}{2} $ at specific angles. In the unit circle, $ \cos\theta = \frac{1}{2} $ at $ \theta = \frac{\pi}{3} $ and $ \theta = \frac{5\pi}{3} $.

To find all solutions, consider the periodic nature of the cosine function. The general solutions are $ \theta = \frac{\pi}{3} + 2\pi n $ and $ \theta = \frac{5\pi}{3} + 2\pi n $, where $ n $ is an integer.

Verify the solutions by substituting back into the original equation to ensure they satisfy it. This confirms the correctness of the general solutions.

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