Math / Trigonometry

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8 9 10 TIME REMAINING 01:58:37 To help solve the trigonometric inequality $2 \sin ^{2}(x)+\cos (2 x)-2<-1$ , Tracey successfully graphed the equations $y=2 \sin ^{2}(x)+\cos (2 x)-2$ and $y=-1$ with her graphing calculator. Which of the following statements is true about the inequality? There is no solution because even though you cannot see it, the graph of the equation $y=2 \sin ^{2}(x)+\cos (2 x)-2$ is above the graph of the equation $y=-1$ for all values of $x$ . There is no solution because the graph of the equation $y=2 \sin ^{2}(x)+\cos (2 x)-2$ is the same as the graph of the equation $y=-1$ , so $y=2 \sin ^{2}(x)+\cos (2 x)-2$ is never below $y=-1$ . There are an infinite number of solutions because even though you cannot see it, the graph of the equation $y=2 \sin ^{2}(x)+\cos (2 x)-2$ is below the graph of the equation $y=-1$ for all values of $x$ . There are an infinite number of solutions because the graph of the equation $y=2 \sin ^{2}(x)+\cos (2 x)-2$ is the same as the graph of the equation $y=-1$ , so $y=2 \sin ^{2}(x)+\cos (2 x)-2$ is never below $y=-1$ .

Studdy Solution

STEP 1

What is this asking? We need to figure out when $2 \sin^2(x) + \cos(2x) - 2$ is less than $-1$ , and what that means about Tracey's graphs. Watch out! Don't just rely on the graphing calculator–a little trig knowledge can go a long way!

STEP 2

1. Simplify the inequality
2. Analyze the simplified inequality

STEP 3

Remember the double-angle formula $\cos(2x) = 1 - 2\sin^2(x)$ .
Let's substitute that into our inequality: $2 \sin^2(x) + (1 - 2\sin^2(x)) - 2 < -1$

STEP 4

Look at that, the $2 \sin^2(x)$ terms add to zero!
So we have: $1 - 2 < -1$ $-1 < -1$

STEP 5

Whoa, $-1 < -1$ is never true!
This tells us something super important: the original inequality, $2 \sin^2(x) + \cos(2x) - 2 < -1$ , is never true for any value of $x$ .

STEP 6

Since the inequality is never true, the graph of $y = 2 \sin^2(x) + \cos(2x) - 2$ is never below the graph of $y = -1$ .
In fact, after simplifying, we found that the left side simplifies to $-1$ .
This means the graph of $y = 2 \sin^2(x) + \cos(2x) - 2$ is actually the same as the graph of $y = -1$ !

STEP 7

There are no solutions because $2 \sin^2(x) + \cos(2x) - 2$ simplifies to $-1$ , which is never less than $-1$ .
So, the correct answer choice is the one that says there's no solution because the graphs are the same, meaning $y = 2 \sin^2(x) + \cos(2x) - 2$ is never below $y = -1$ .

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Studdy Solution

STEP 1

What is this asking? We need to figure out when $2 \sin^2(x) + \cos(2x) - 2$ is less than $-1$ , and what that means about Tracey's graphs. Watch out! Don't just rely on the graphing calculator–a little trig knowledge can go a long way!

STEP 2

1. Simplify the inequality
2. Analyze the simplified inequality

STEP 3

Remember the double-angle formula $\cos(2x) = 1 - 2\sin^2(x)$ .
Let's substitute that into our inequality: $2 \sin^2(x) + (1 - 2\sin^2(x)) - 2 < -1$

STEP 4

Look at that, the $2 \sin^2(x)$ terms add to zero!
So we have: $1 - 2 < -1$ $-1 < -1$

STEP 5

Whoa, $-1 < -1$ is never true!
This tells us something super important: the original inequality, $2 \sin^2(x) + \cos(2x) - 2 < -1$ , is never true for any value of $x$ .

STEP 6

STEP 7

There are no solutions because $2 \sin^2(x) + \cos(2x) - 2$ simplifies to $-1$ , which is never less than $-1$ .
So, the correct answer choice is the one that says there's no solution because the graphs are the same, meaning $y = 2 \sin^2(x) + \cos(2x) - 2$ is never below $y = -1$ .

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Studdy Solution

STEP 1

What is this asking? We need to figure out when 2sin⁡2(x)+cos⁡(2x)−22 \sin^2(x) + \cos(2x) - 22sin2(x)+cos(2x)−2 is less than −1-1−1, and what that means about Tracey's graphs. Watch out! Don't just rely on the graphing calculator–a little trig knowledge can go a long way!

STEP 2

1. Simplify the inequality2. Analyze the simplified inequality

STEP 3

Remember the double-angle formula cos⁡(2x)=1−2sin⁡2(x)\cos(2x) = 1 - 2\sin^2(x)cos(2x)=1−2sin2(x).Let's **substitute** that into our inequality: 2sin⁡2(x)+(1−2sin⁡2(x))−2<−12 \sin^2(x) + (1 - 2\sin^2(x)) - 2 < -12sin2(x)+(1−2sin2(x))−2<−1

STEP 4

Look at that, the 2sin⁡2(x)2 \sin^2(x)2sin2(x) terms add to zero!So we have: 1−2<−11 - 2 < -11−2<−1 −1<−1-1 < -1−1<−1

STEP 5

Whoa, −1<−1-1 < -1−1<−1 is *never* true!This tells us something super important: the original inequality, 2sin⁡2(x)+cos⁡(2x)−2<−12 \sin^2(x) + \cos(2x) - 2 < -12sin2(x)+cos(2x)−2<−1, is **never true** for *any* value of xxx.

STEP 6

STEP 7

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Studdy solves anything!

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to figure out when $2 \sin^2(x) + \cos(2x) - 2$ is less than $-1$ , and what that means about Tracey's graphs. Watch out! Don't just rely on the graphing calculator–a little trig knowledge can go a long way!

1. Simplify the inequality
2. Analyze the simplified inequality

Remember the double-angle formula $\cos(2x) = 1 - 2\sin^2(x)$ .
Let's substitute that into our inequality: $2 \sin^2(x) + (1 - 2\sin^2(x)) - 2 < -1$

Look at that, the $2 \sin^2(x)$ terms add to zero!
So we have: $1 - 2 < -1$ $-1 < -1$

Whoa, $-1 < -1$ is never true!
This tells us something super important: the original inequality, $2 \sin^2(x) + \cos(2x) - 2 < -1$ , is never true for any value of $x$ .