Number of solutions of 3sin2x+cos2x+3sin⁡x+cos⁡x+1=3sin⁡xcos⁡x in [0 10π]My Try:The given equation is2+2sin2x+2sin⁡(x+π6)=32sin⁡(2x)⇒3+2sin⁡(x+π6)=32sin⁡(2x)+cos⁡(2x)

We can rewrite the equation as (3sin⁡x−cos⁡x)2+(1+3sin⁡x)(1+cos⁡x)=0Let a=1+3sin⁡x and b=1+cos⁡xThen the equation is (a−b)2+ab=0⇒a2−ab+b2=0But a2−ab+b2=0⇔a=0,b=0 which is not possible.Hence the equation has no solutions.

Expert Answer to "Number of solutions of 3sin2x+cos2x+3sin⁡x+cos⁡x+1=3sin⁡xcos⁡x in [0 ..."

Sydney Stanley

Answered question

2022-04-22

Number of solutions of

3 \sin^{2} x + \cos^{2} x + \sqrt{3} \sin x + \cos x + 1 = \sqrt{3} \sin x \cos x

[0 10 π]

My Try:
The given equation is

2 + 2 \sin^{2} x + 2 \sin (x + \frac{π}{6}) = \frac{\sqrt{3}}{2} \sin (2 x)

\Rightarrow

3 + 2 \sin (x + \frac{π}{6}) = \frac{\sqrt{3}}{2} \sin (2 x) + \cos (2 x)

Answer & Explanation

Barbara Navarro

Beginner2022-04-23Added 18 answers

We can rewrite the equation as

{(\sqrt{3} \sin x - \cos x)}^{2} + (1 + \sqrt{3} \sin x) (1 + \cos x) = 0

Let

a = 1 + \sqrt{3} \sin x

and

b = 1 + \cos x

Then the equation is

{(a - b)}^{2} + a b = 0 \Rightarrow a^{2} - a b + b^{2} = 0

But

a^{2} - a b + b^{2} = 0 \Leftrightarrow a = 0, b = 0

which is not possible.
Hence the equation has no solutions.

Genesis Reilly

Beginner2022-04-24Added 12 answers

Let

c = \cos (x)

and

s = \sin (x)

, solve for c:

3 s^{2} + c^{2} + \sqrt{3} s + c + 1 = \sqrt{3} c s

c^{2} + (1 - \sqrt{3} s) c + 3 s^{2} + \sqrt{3} s + 1 = 0

c = \frac{\sqrt{3} s - 1 \pm \sqrt{{(1 - \sqrt{3} s)}^{2} - 4 (3 s^{2} + \sqrt{3} s + 1)}}{2}

Normally here you'd check the range on the discriminate, but the discriminate is a perfect square, so:

c = \frac{\sqrt{3} s - 1 \pm \sqrt{- 9 s^{2} - 6 \sqrt{3} s - 3}}{2}

c = \frac{\sqrt{3} s - 1 \pm i \sqrt{{(3 s + \sqrt{3})}^{2}}}{2}

c = \frac{\sqrt{3} s - 1 \pm i | 3 s + \sqrt{3} |}{2}

So if s is real then c is complex, so no solutions.

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