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3 years ago

Simplify the expression. Write your answer using decimals.

Mathematics

1 answer:

Andrei [34K]3 years ago

7 0

Answer:

−2.5d−0.7

Step-by-step explanation:

1 Expand by distributing terms.

-(2.5d+5)+4.3

2 Remove parentheses.

-2.5d-5+4.3

3 Collect like terms.

-2.5d+(-5+4.3)

4 Simplify.

-2.5d-0.7

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3 years ago

Solve sinX+1=cos2x for interval of more or equal to 0 and less than 2pi

Igoryamba

Answer:

Question 1: $\sin(x)+1=\cos^2(x)$

Answer to Question 1: $x=0, \pi \frac{3\pi}{2}$

Question 2: $\sin(x)+1=\cos(2x)$

Answer to Question 2: $0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}$

Question:

I will answer the following two questions.

Condition: $0\le x$

Question 1: $\sin(x)+1=\cos^2(x)$

Question 2: $\sin(x)+1=\cos(2x)$

Step-by-step explanation:

Question 1: $\sin(x)+1=\cos^2(x)$

Question 2: $\sin(x)+1=\cos(2x)$

Question 1:

$\sin(x)+1=\cos^2(x)$

I will use a Pythagorean Identity so that the equation is in terms of just one trig function, $\sin(x)$ .

Recall $\sin^2(x)+\cos^2(x)=1$ .

This implies that $\cos^2(x)=1-\sin^2(x)$ . To get this equation from the one above I just subtracted $\sin^2(x)$ on both sides.

So the equation we are starting with is:

$\sin(x)+1=\cos^2(x)$

I'm going to rewrite this with the Pythagorean Identity I just mentioned above:

$\sin(x)+1=1-\sin^2(x)$

This looks like a quadratic equation in terms of the variable: $\sin(x)$ .

I'm going to get everything to one side so one side is 0.

Subtracting 1 on both sides gives:

$\sin(x)+1-1=1-\sin^2(x)-1$

$\sin(x)+0=1-1-\sin^2(x)$

$\sin(x)=0-\sin^2(x)$

$\sin(x)=-\sin^2(x)$

Add $\sin^2(x)$ on both sides:

$\sin(x)+\sin^2(x)=-\sin^2(x)+\sin^2(x)$

$\sin(x)+\sin^2(x)=0$

Now the left hand side contains terms that have a common factor of $\sin(x)$ so I'm going to factor that out giving me:

$\sin(x)[1+\sin(x)]=0$

Now this equations implies the following:

$\sin(x)=0$ or $1+\sin(x)=0$

$\sin(x)=0$ when the $y$ -coordinate on the unit circle is 0. This happens at $0$ , $\pi$ , or also at $2\pi$ . We do not want to include $2\pi$ because of the given restriction $0\le x$ .