Question

How do you find the exact value of `2cos^2theta-3sintheta=0` in the interval `0<=theta<360`?

Answer 1

`theta=30^@" or "theta=150^@`

Explanation:

`"using the "color(blue)"trigonometric identity"`

`•color(white)(x)sin^2x+cos^2x=1`

`rArrcos^2x=1-sin^2x`

`rArr2(1-sin^2theta)-3sintheta=0`

`rArr2-2sin^2theta-3sintheta=0`

`rArr2sin^2theta+3sintheta-2=0`

`"we have a quadratic in sine which factors as"`

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

`"equate each factor to zero and solve for "theta`

`sintheta+2=0larrcolor(blue)"has no solution"`

`"since "-1<=sintheta<=1`

`2sintheta-1=0rArrsintheta=1/2`

`"since "sintheta>0" then "theta" in first/second quadrant"`

`theta=sin^-1(1/2)=30^@larrcolor(red)"first quadrant"`

`"or "theta=(180-30)^@=150^@larrcolor(red)"second quad"`

`rArrtheta=30^@" or "theta=150^@to(0<=theta<=360)`