Question

solve cosx - sqrt3sinx = 2
a. x = 60°
b. x = 300°
c. x = 60° , 240°
d. x = 0°, 60° , 240°

Answer 1

\(\color{#000000 }{ \displaystyle \cos x - \sqrt{~3~}\sin x =2 }\)

You know that:
\(\color{#000000 }{ \displaystyle \cos^2x + \sin^2x=1 }\)
\(\color{#000000 }{ \displaystyle \cos^2x =1- \sin^2x }\)
\(\color{#000000 }{ \displaystyle \cos x =\sqrt{1- \sin^2x} }\)

So,

\(\color{#000000 }{ \displaystyle \sqrt{1- \sin^2x}- \sqrt{~3~}\sin x =2 }\)

\(\color{#000000 }{ \displaystyle \sqrt{1- \sin^2x} =2+\sqrt{~3~}\sin x }\)

\(\color{#000000 }{ \displaystyle (\sqrt{1- \sin^2x})^2 =(2+\sqrt{~3~}\sin x)^2 }\)

\(\color{#000000 }{ \displaystyle 1- \sin^2x=(2+\sqrt{~3~}\sin x)^2 }\)

Answer 2

simplify that, and then let sin(x)=a

Answer 3

Then, there is another rule:
\(\color{#000000 }{ \displaystyle(a+b)^2=a^2+2ab+b^2 }\)

And in this case,
\(\color{#000000 }{ \displaystyle (2+\sqrt{~3~}\sin x)^2=2^2+2\cdot 2(\sqrt{~3~}\sin x)+(\sqrt{~3~}\sin x)^2 }\)

Answer 4

thank you!