Trigonometric functions (1 Viewer)

[Abide]

Hey, could anyone please help me do this question??

solve the following equation for 0 <\ x <\ 2pi

3 - cos^2 - 3sinx = 0

?

Ta..

 

[Antisocial]

Tez. You could've at least asked me.

Remember your trigonometric identities (sinx + cosx = 1).
In this question, you need to substitute cosx with 1 - sinx.

So it's...
(Remembering the brackets) -
3 - (1 - sinx) - 3sinx = 0
(Simplify and rearrange) -
sinx - 3sinx + 2 = 0
(Solve as per normal quadratic equation) -
(sinx - 2)(sinx - 1) = 0

And it's sinx = 2 or sinx = 1.
It cannot be sinx = 2, because if you graph sinx, the maximum and minimum points are 1 and -1, respectively. So...

sinx = 1

Therefore, x = pi/2 (which is equivalent to 90).