# solve for theta, to nearest tenth of a degree, in interval 0 degrees<=theta<=360 degrees for equation: 3sec^2(theta) - 5tan(theta)=1 You need to remember that `1/(cos^2 theta)=sec^2theta`  and you need to substitute `1 + tan^2theta for 1/(cos^2 theta)`  such that:

`3(1 + tan^2 theta) - 5tan(theta)=1`

You need to open the brackets such that:

`3 + 3tan^2 theta - 5tan theta = 1`

You need to move all terms to one...

You need to remember that `1/(cos^2 theta)=sec^2theta`  and you need to substitute `1 + tan^2theta for 1/(cos^2 theta)`  such that:

`3(1 + tan^2 theta) - 5tan(theta)=1`

You need to open the brackets such that:

`3 + 3tan^2 theta - 5tan theta = 1`

You need to move all terms to one side such that:

`3tan^2 theta - 5tan theta + 3 - 1 = 0`

You should come up with the substitution `tan theta = y`  such that:

`3y^2 - 5y + 2 =`  0

You need to use quadratic formula to find `y_1`  and `y_2`  such that:

`y_(1,2) = (5+-sqrt(25 - 24))/6 =gt y_(1,2) = (5+-1)/6`

`y_1 = 1 ; y_2 = 2/3`

You need to find theta, hence you need to solve the equations `tan theta = 1`  and tan `theta = 2/3` .

You need to remember that tangent function is positive in quadrants I and III.

`tan theta = 1 =gt theta = pi/4` `= 45^o`

`theta = pi + pi/4 =gt theta = 5pi/4` `= 225^o`

`tan theta = 2/3 =gt theta = tan^(-1) (2/3)`

`theta~~33^o 42' `

`theta~~213^o 42'`

Hence, evaluating solutions to  given equation yields `theta=33^o 42'` , `theta = 45^o, theta = 213^o 42', theta = 225^o`.

Approved by eNotes Editorial Team `3sec^2theta-5tantheta=1=>`

`3/[cos^2theta]-5sintheta/costheta=1=>`

`3/[cos^2theta]-[5sinthetacostheta]/[cos^2theta]=1=>`

`3-5sinthetacostheta=cos^2theta`

`cos^2theta+5sinthetacostheta-3=0`

let `x=sintheta`

then `cos^2theta=1-x^2`

Hence we can rewrite the equation

`1-x^2-5xsqrt(1-x^2)-3=0=>`

`-2-x^2=5xsqrt(1-x^2)=>` square both sides you get

`(-2-x^2)^2=25x^2(1-x^2)=>`

`x^4+4x^2+4=25x^2-25x^4=>`

`26x^4-21x^2+4=0=>` we can solve the following by factoring or using the quadratic formula

`(2x^2-1)(13x^2-4)=0=>`

`2x^2-1=0 =>x^2=1/2=>x=+-1/sqrt2`

or

`13x^2-4=0=>x^2=4/13=>x=+-2/sqrt(13)`

so our four solution are `sqrt(2)/2, -sqrt(2)/2, 2/sqrt(13),-2/sqrt(13)`

Thus

`sintheta=sqrt2/2=>theta=45 or 180-45=135`

`sintheta=-sqrt(2)/2=>theta=45+180=225 or 360-45=315`

`sintheta=2/sqrt13=>theta=33.7 or 180-33.7=146.3`

`sintheta=-2/sqrt13=>theta=360-33.7=326.3 or theta=180+33.7=213.7`

So theta can be 45, 135, 225, 315, 33.7, 146.3, 326.3, or 213.7

An important addition, because we solved for sin^2, we have several extraneous solutions. If we try to plug in the above values in the original equation, we will find that 135, 315, 146.3, and 326.3 are not solutions.

So our solution set is `{45,225,33.7,213.7}`

Approved by eNotes Editorial Team