# Please explain how to get the answer to this equation if `1+sintheta` =`2cos^(2)theta` , `0<=theta<=2pi` , *what I need is an explanation about why this is the answer! To solve for...

Please explain how to get the answer to this equation if `1+sintheta` =`2cos^(2)theta` , `0<=theta<=2pi` ,

*what I need is an explanation about why this is the answer!

To solve for `theta` in the equation `1+sintheta=2cos^2theta` in the interval`0<=theta<=2pi`, we substitute for`cos^(2)theta`from the identity `sin^(2)+cos^2theta=1` ,i.e., `cos^(2)theta=1-sin^2theta`

Hence, `1+sintheta` `=2cos^(2)theta `

`1+sintheta=2(1-sin^2theta)=2-2sin^2theta`

`2sin^2theta+sintheta-1=0.`

`(2sintheta-1)(sintheta+1)=0`

`2sintheta=1 rArrsintheta=1/2`

`sintheta=-1rArrtheta=(3pi)/(2)`

`theta=pi/6, (5pi)/(6) Therefore, theta=pi/6 , (5pi)/(6) , (3pi)/(2)`

` ``0<=theta<=2pi`

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Your explanation of the rules regarding identities is good. Being a "rule" make sure learn it as you will use it considerably. Now that we know that `cos^2 theta = 1` we substitute (in terms of what you did) but we have `2cos^2 theta` hence:

`1 + sin theta=2cos^2 theta` becomes

`1+sin theta= 2(1-sin^2theta)`

Now remove the brackets:

`therefore 1 +sin theta= 2-2sin^2 theta`

Now using the rules you learnt when doing algebra, get all your terms to one side:

`therefore 1 + sin theta - 2 + 2sin ^2theta=0` (note the symbols have changed when moved from RHS- right hand side- to LHS )

Now rearrange and add like terms:

`therefore 2 sin^2 theta + sin theta -1 = 0` (note we had +1 and -2=-1)

Again using the same rules as algebra - not the three tems - find the factors of the first term `(2sin theta times sin theta)` and the factors of the third term (`1 times -1)` and arrange the factors to ensure a middle term of `+sin theta`

`therefore (2sin theta-1)(sin theta+1)=0`

Make each factor equal to zero:

`2 sin theta-1=0` and `sin theta +1=0`

`therefore 2sin theta = 1` and `sin theta=-1`

`therefore sin theta = 1/2` `= 30 `deg and `sin theta= -1 = 270` deg

We know that `pi= 180` and `180 divide 6= 30`

`therefore theta=pi/6` .

Furthermore, `sin 30 = 1/2` at 150 deg (quad II where sin is positive) as `sin (180-30)=sin150`

and `5/6 times 180 = 150`

`therefore(5pi)/6= theta` (remembering that `pi=180` deg

and `270=3/2 times 180` `therefore theta=(3pi)/2`