# solve for theta, to nearest tenth of a degree, in interval 0 degrees<=theta<=360 degrees for equation: 3sec^2(theta) - 5tan(theta)=1

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`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}`

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`.**