Question

Question #96293

Answer 1

On the domain, `{x in R|0<=x<=2pi}`

`x=(pi)/6, x=(5pi)/6 ,x=(7pi)/6, x=(11pi)/6`

Or, in degrees;

On the domain, `{x in R|0^@<=x<=360^@}`

`x=30^@, x=150^@ ,x=210^@, x=330^@`

Explanation:

You are going to need to use the pythagorean identity (`sin^2x+cos^2x=1`) to solve this problem.

`7 sin^2x+3 cos^2x=4`

As, `sin^2x+cos^2x=1` then; `cos^2x=1-sin^2x` which we can use to substitute into the given equation.

`7 sin^2x+3 (color(red)(1-sin^2x))=4`

`7sin^2x+3-3sin^2x=4`

Combining like terms;

`4sin^2x+3=4`

Moving all of the information to the left of the equality yields;

`4sin^2x-1=0`

The next clever piece of maths comes because you will need to factorise this expression. So what we do is substitute a variable in for the trigonometric ratio.

i.e. let `u=sinx`

`:. 4sin^2x-1=4u^2-1=0`

Which we can factorise using the difference of perfect squares.

`4u^2-1=(2u+1)(2u-1)=0`

By the null factor law we can conclude that `u=+-1/2`

`:. sinx=+-1/2`

So, on the domain, `{x in R|0<=x<=2pi}`

`x=(pi)/6, x=(5pi)/6 ,x=(7pi)/6, x=(11pi)/6`

Or, in degrees;

On the domain, `{x in R|0^@<=x<=360^@}`

`x=30^@, x=150^@ ,x=210^@, x=330^@`

I hope that helps :)