Number of principle solutions of the equation

[DaalChawal]

Find number of principal solutions of the equation
$sinx + cosx + sin2x + cos2x + sin3x = -1 $
Please help anyone wihout using desmos

 

[jvicens]

There are a few different ways to approach this problem without using Desmos. One method is to rewrite the equation using trigonometric identities to simplify it. For example, we can rewrite the equation as:

$sinx + cosx + 2sinxcosx + 2cos^2x - 1 = -1$

Then, we can use the double angle formula for cosine to rewrite the equation as:

$sinx + cosx + 2sinxcosx + 1 - 2sin^2x = -1$

Next, we can rearrange the terms to get:

$2sinxcosx - 2sin^2x = -1 - sinx - cosx$

Now, we can use the Pythagorean identity for sine to rewrite the equation as:

$2sinxcosx - 2(1-cos^2x) = -1 - sinx - cosx$

After simplifying, we get:

$2cos^2x + 2sinx - 1 = 0$

This is now a quadratic equation in terms of cosx. We can use the quadratic formula to solve for cosx:

$cosx = \frac{-2 \pm \sqrt{4 - 4(2)(-1)}}{2(2)}$

Simplifying further, we get:

$cosx = \frac{-2 \pm \sqrt{12}}{4}$

$cosx = \frac{-2 \pm 2\sqrt{3}}{4}$

$cosx = \frac{-1 \pm \sqrt{3}}{2}$

Now, we can plug these values back into the original equation to solve for x. We get two possible solutions:

$x = \frac{\pi}{3} + 2\pi n$ or $x = \frac{5\pi}{6} + 2\pi n$ where n is any integer.

Therefore, there are two principal solutions for the given equation.