# What is the answer for question 1) ? http://postimg.org/image/con6jawnr/

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

Solve `-cos^2x=-cosx` on the interval `0<=x<=360^@` :

Note: do not divide both sides by cosx; this will result in a lost root. Add `cos^2x` to both sides to get:

`cos^2x-cosx=0`

`cosx(cosx-1)=0`

Either cosx=0 ==> `x=90^@,270^@`

or `cosx-1=0 ==> cosx=1 ==> x=0,360^@`

-------------------------------------------------------------

The solutions are `x=0,90^@,270^@,360^@`

---------------------------------------------------------------

The graph of `y=-cos^2x` in black, and y=-cosx in red:

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

We are asked to answer 1 specific question per post. This allows users to search for questions easier.

Solve `6sin^2x-11sinx+4=0` on the interval `0<=x<=360^@` . (This is quadratic in sinx.)

Factor the left side:

`(3sinx-4)(2sinx-1)=0`

By the zero product property at least one of the factors must be zero:

(1) `2sinx-1=0 ==gt sinx=1/2 ==gt x=30^@,x=150^@`

(2) `3sinx-4=0 ==gt sinx=4/3` which is impossible as the range for the sine is -1 to 1.

------------------------------------------------------------------

` x=30^@,x=150^@` are the solutions

------------------------------------------------------------------