Math  /  Trigonometry

QuestionSolve the following equations
1. cos(4x)8=0\cos (4 x)-8=0 for the four smallest positive solutions.
2. cos(x)(2sin(x)+1)=0\cos (x) \cdot(2 \sin (x)+1)=0 on the interval [0,2π][0,2 \pi]
3. 2cos2(x)+3cos(x)+1=02 \cos ^{2}(x)+3 \cos (x)+1=0 on the interval [0,2π][0,2 \pi]

Studdy Solution

STEP 1

What is this asking? We need to find the values of xx that make three different trigonometric equations true, with some restrictions on how big xx can be. Watch out! Remember that trig functions can have multiple solutions, and we need to make sure we find all of them within the given ranges!
Also, be careful with those squared trig functions – they can be a bit tricky!

STEP 2

1. Solve for cos(4x)
2. Find general solutions for 4x
3. Find specific solutions for x
4. Solve equation with factored form
5. Solve quadratic equation in cosine

STEP 3

We're given cos(4x)8=0\cos(4x) - 8 = 0.
Let's **isolate** the cos(4x)\cos(4x) by adding **8** to both sides.
This gives us cos(4x)=8\cos(4x) = 8.

STEP 4

Uh oh!
The cosine function only outputs values between **-1** and **1**.
Since **8** is outside of this range, there are **no solutions** for xx in this case.
Sometimes, no solution *is* the solution!

STEP 5

Our equation is cos(x)(2sin(x)+1)=0\cos(x) \cdot (2\sin(x) + 1) = 0.
This means either cos(x)=0\cos(x) = 0 or 2sin(x)+1=02\sin(x) + 1 = 0.
Let's tackle these one at a time!

STEP 6

If 2sin(x)+1=02\sin(x) + 1 = 0, then we subtract **1** from both sides and divide by **2** to get sin(x)=12\sin(x) = -\frac{1}{2}.

STEP 7

For cos(x)=0\cos(x) = 0, we know xx can be π2\frac{\pi}{2} or 3π2\frac{3\pi}{2} plus any multiple of 2π2\pi.
For sin(x)=12\sin(x) = -\frac{1}{2}, xx can be 7π6\frac{7\pi}{6} or 11π6\frac{11\pi}{6} plus any multiple of 2π2\pi.

STEP 8

We're looking for solutions in the interval [0,2π][0, 2\pi].
Looking at our general solutions from the previous step, we see that π2\frac{\pi}{2}, 3π2\frac{3\pi}{2}, 7π6\frac{7\pi}{6}, and 11π6\frac{11\pi}{6} all fall within this interval.
Any other solutions would be outside of this range.

STEP 9

We have 2cos2(x)+3cos(x)+1=02\cos^2(x) + 3\cos(x) + 1 = 0.
This looks like a quadratic equation!
Let's replace cos(x)\cos(x) with uu to get 2u2+3u+1=02u^2 + 3u + 1 = 0.

STEP 10

We can **factor** this quadratic as (2u+1)(u+1)=0(2u + 1)(u + 1) = 0.

STEP 11

This means 2u+1=02u + 1 = 0 or u+1=0u + 1 = 0, so u=12u = -\frac{1}{2} or u=1u = -1.

STEP 12

Since u=cos(x)u = \cos(x), we have cos(x)=12\cos(x) = -\frac{1}{2} or cos(x)=1\cos(x) = -1.

STEP 13

For cos(x)=12\cos(x) = -\frac{1}{2}, xx can be 2π3\frac{2\pi}{3} or 4π3\frac{4\pi}{3} in the interval [0,2π][0, 2\pi].
For cos(x)=1\cos(x) = -1, xx can be π\pi in the given interval.

STEP 14

1. No solutions for cos(4x)8=0\cos(4x) - 8 = 0.
2. For cos(x)(2sin(x)+1)=0\cos(x) \cdot (2\sin(x) + 1) = 0 on [0,2π][0, 2\pi], xx can be π2\frac{\pi}{2}, 3π2\frac{3\pi}{2}, 7π6\frac{7\pi}{6}, or 11π6\frac{11\pi}{6}.
3. For 2cos2(x)+3cos(x)+1=02\cos^2(x) + 3\cos(x) + 1 = 0 on [0,2π][0, 2\pi], xx can be 2π3\frac{2\pi}{3}, 4π3\frac{4\pi}{3}, or π\pi.

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