Question

1+cos(pi/6-x)=sin^2x+0,5sinx ХЕЛП

Answer 1

Ответ:

Объяснение:

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$\left [\begin{array}{lcl} {{x=\frac{\pi }{2} +\pi n,~n\in\mathbb{Z}, } \\\\ {x=\pm\frac{5\pi }{6} }+2\pi k,~k\in\mathbb{Z}. } \end{array} \right.$

Answer 2

Ответ:

Объяснение:

1 + cos Π/6*cos x + sin Π/6*sin x = sin^2 x + 0,5*sin x

1 + √3/2*cos x + 0,5*sin x = sin^2 x + 0,5*sin x

1 + √3/2*cos x = sin^2 x

1 - sin^2 x + √3/2*cos x = 0

cos^2 x + √3/2*cos x = 0

cos x*(cos x + √3/2) = 0

1) cos x = 0

x1 = Π/2 + Π*k; k € Z

2) cos x = -√3/2

x2 = 5Π/6 + 2Π*n; n € Z

x3 = 7Π/6 + 2Π*n; n € Z

Answer 3

To solve the equation $1 + \cos\left(\frac{\pi}{6} - x\right) = \sin^2x + 0.5\sinx$, we can use trigonometric identities and algebraic manipulations. Let's go step by step:

Step 1: Simplify the left-hand side (LHS): We'll use the trigonometric identity $\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta$.

$1 + \cos\left(\frac{\pi}{6} - x\right) = 1 + \left(\cos\frac{\pi}{6} \cos x + \sin\frac{\pi}{6} \sin x\right)$

Now, we know that $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \frac{1}{2}$:

$1 + \left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right)$

Step 2: Simplify the right-hand side (RHS): The RHS is $\sin^2x + 0.5\sinx$.

Step 3: Rewrite $\sin^2x$ in terms of $\cos x$: We'll use the identity $\sin^2x = 1 - \cos^2x$.

$\sin^2x = 1 - \cos^2x$

Step 4: Substitute $\sin^2x$ in the RHS: $1 - \cos^2x + 0.5\sinx$

Step 5: Rewrite $0.5\sinx$ in terms of $\cos x$: We know that $\sinx = 2\sin\frac{x}{2} \cos\frac{x}{2}$.

$0.5\sinx = \sin\frac{x}{2} \cos\frac{x}{2}$

Step 6: Substitute $0.5\sinx$ in the RHS: $1 - \cos^2x + \sin\frac{x}{2} \cos\frac{x}{2}$

Now, the equation becomes: $1 + \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x = 1 - \cos^2x + \sin\frac{x}{2} \cos\frac{x}{2}$

Step 7: Move all terms to one side to set the equation to zero: $\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x - \cos^2x + \sin\frac{x}{2} \cos\frac{x}{2} = 0$

Step 8: Factor out common terms from the trigonometric terms: $\cos x \left(\frac{\sqrt{3}}{2} + \cos x + \frac{1}{2} \sin\frac{x}{2}\right) + \sin\frac{x}{2} \cos\frac{x}{2} = 0$