Question

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

Answer 1

thats the first page and the second page

the answer is x {-π/2+1+2kπ}

x={-1/3+π/6+2kπ}

i hope it helps:)

Answer 2

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

Answer to Question 1:
`x=0, \pi (3\pi)/(2)`

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

Answer to Question 2:
`0,\pi,(7\pi)/(6), (11\pi)/(6)`

Question:

I will answer the following two questions.

Condition:
`0\le x <2\pi`

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

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

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 <2\pi`.

We must also solve
`1+\sin(x)=0`.

Subtract 1 on both sides:

`\sin(x)=-1`

We are looking for when the
`y`-coordinate is -1.

This happens at
`(3\pi)/(2)` on the unit circle.

So the solutions to question 1 are
`0,\pi,(3\pi)/(2)`.

Question 2:

`\sin(x)+1=\cos(2x)`

So the objective at the beginning is pretty much the same. We want the same trig function.

`\cos(2x)=\cos^2(x)-\sin^2(x)` by double able identity for cosine.

`\cos(2x)=(1-\sin^2(x))-\sin^2(x)` by Pythagorean Identity.

`\cos(2x)=1-2\sin^2(x)` (simplifying the previous equation).

So let's again write in terms of the variable
`\sin(x)`.

`\sin(x)+1=\cos(2x)`

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

Subtract 1 on both sides:

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

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

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

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

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

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

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

Now on the left hand side there are two terms with a common factor of
`\sin(x)` so let's factor that out:

`\sin(x)[1+2\sin(x)]=0`

This implies
`\sin(x)=0` or
`1+2\sin(x)=0`.

The first equation was already solved in question 1. It was just at
`x=0`.

Let's look at the other equation:
`1+2\sin(x)=0`.

Subtract 1 on both sides:

`2\sin(x)=-1`

Divide both sides by 2:

`\sin(x)=(-1)/(2)`

We are looking for when the
`y`-coordinate on the unit circle is
`(-1)/(2)`.

This happens at
`(7\pi)/(6)` or also at
`(11\pi)/(6)`.

So the solutions for this question 2 is
`0,\pi,(7\pi)/(6), (11\pi)/(6)`.