Math / Trigonometry

Question13. Determine approximate solutions for $\cos ^{2} x-0.49=0$ in the interval $x \in[0,2 \pi]$ , to the nearest hundredth of a radian.

Studdy Solution

STEP 1

What is this asking? Find all the angles $x$ between $0$ and $2\pi$ where $\cos^2 x - 0.49 = 0$ , rounded to the nearest hundredth. Watch out! There are going to be multiple solutions, so don't stop after finding just one!
Also, remember that $\cos^2 x$ means $(\cos x)^2$ , not $\cos(\cos(x))$ !

STEP 2

1. Isolate the cosine
2. Find the principal angles
3. Find all the angles

STEP 3

We want to get $\cos x$ by itself.
First, let's add $0.49$ to both sides of the equation to isolate the cosine term: $\cos^2 x - 0.49 + 0.49 = 0 + 0.49$ $\cos^2 x = 0.49$

STEP 4

Now, take the square root of both sides.
Remember, when we take the square root, we get both positive and negative solutions! $\sqrt{\cos^2 x} = \pm\sqrt{0.49}$ $\cos x = \pm 0.7$

STEP 5

Let's find the principal angle for $\cos x = 0.7$ .
Using a calculator, we find: $x = \arccos(0.7) \approx 0.7954$ So, one solution is approximately $x \approx \mathbf{0.80}$ radians.

STEP 6

Now, let's find the principal angle for $\cos x = -0.7$ .
Using a calculator: $x = \arccos(-0.7) \approx 2.3462$ So, another solution is approximately $x \approx \mathbf{2.35}$ radians.

STEP 7

Remember the symmetry of the cosine function!
If $\cos x = 0.7$ , then $\cos(2\pi - x)$ is also equal to $0.7$ .
Similarly, if $\cos x = -0.7$ , then $\cos(2\pi - x)$ is also equal to $-0.7$ .

STEP 8

Using our principal angles, we can find the other solutions in the interval $[0, 2\pi]$ . For $x \approx 0.80$ , the other solution is $2\pi - 0.80 \approx 2 \cdot 3.1416 - 0.80 \approx 6.2832 - 0.80 \approx \mathbf{5.48}$ . For $x \approx 2.35$ , the other solution is $2\pi - 2.35 \approx 6.2832 - 2.35 \approx \mathbf{3.93}$ .

STEP 9

The approximate solutions for $\cos^2 x - 0.49 = 0$ in the interval $x \in [0, 2\pi]$ are $x \approx \mathbf{0.80}$ , $x \approx \mathbf{2.35}$ , $x \approx \mathbf{3.93}$ , and $x \approx \mathbf{5.48}$ radians.

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