Question

Solve cos x - sin x tan x + 1 = 0 for 0°≤x≤360° ? need the solution asap, thank you.

Answer 1

`" The Soln. Set."sub[0^@,360^@]={60^@, 180^@, 300^@}`.

Explanation:

`cosx-sinxtanx+1=0`.

`:. cosx-sinx*sinx/cosx+1+0`.

`:. (cos^2x-sin^2x+cosx)/cosx=0`.

`:. (cos^2x-sin^2x)+cosx=0`.

`:. cos2x+cosx=0`.

`:. 2cos((2x+x)/2)cos((2x-x)/2)=0`.

`:. cos(3*x/2)cos(x/2)=0`.

Since, `cos3theta=4cos^3theta-3costheta`, we have,

`:. {4cos^3(x/2)-3cos(x/2)}cos(x/2)=0`.

`:. cos^2(x/2){4cos^2(x/2)-3}=0`.

`:. cos(x/2)=0 or 4cos^2(x/2)=3`.

`:. cos(x/2)=0 or cos(x/2)=+-sqrt3/2`.

Because, `0^@ le x le 360^@, 0^@ le x/2 le 180^@`.

`:. cos(x/2)=0 rArr x/2=90^@ rArr x=180^@`.

Further, `cos(x/2)=sqrt3/2 rArr x/2=30^@ rArr x=60^@, &,`

`cos(x/2)=-sqrt3/2 rArr x/2=180^@-30^@=150^@. :. x=300^@`.

`:." The Soln. Set."sub[0^@,360^@]={60^@, 180^@, 300^@}`.

Feel the Joy of Maths.!

Answer 2

`pi/3, pi, (5pi)/3`

Explanation:

cos x - sin x.tan x = - 1
`cos^2 x - sin^2 x = - cos x`
`cos 2x = - cos x = cos (x + pi)`
Unit circle and property of cos function give -->
`2x = +- (x + pi)`
a. `2x = x + pi`
`x = pi`
b. `2x = - x - pi`
`3x = - pi`, and `3x = pi`
1. `3x = - pi` --> `x = - pi/3`, or `x = (5pi)/3` (co-terminal).
2. `3x = pi` --> `x = pi/3`
Check by calculator
`x = (5pi)/3 = 300` --> cos 300 = 0.5 --> sin 300 = -0.87 -->
tan 300 = - 1.732
cos x - sin x.tan x = 0.5 - (-0.87)(-1,732) = 0.5 - 1.5 = - 1. Proved