Gỉai pt 6cos^2x + 5sinx -7 =0

Đáp án: Simplifying

6cos2x + 5sinx + -7 = 0

Reorder the terms:

-7 + 6cos2x + 5insx = 0

Solving

-7 + 6cos2x + 5insx = 0

Solving for variable 'c'.

Move all terms containing c to the left, all other terms to the right.

Add '7' to each side of the equation.

-7 + 6cos2x + 7 + 5insx = 0 + 7

Reorder the terms:

-7 + 7 + 6cos2x + 5insx = 0 + 7

Combine like terms: -7 + 7 = 0

0 + 6cos2x + 5insx = 0 + 7

6cos2x + 5insx = 0 + 7

Combine like terms: 0 + 7 = 7

6cos2x + 5insx = 7

Add '-5insx' to each side of the equation.

6cos2x + 5insx + -5insx = 7 + -5insx

Combine like terms: 5insx + -5insx = 0

6cos2x + 0 = 7 + -5insx

6cos2x = 7 + -5insx

Divide each side by '6os2x'.

c = 1.166666667o-1s-2x-1 + -0.8333333333ino-1s-1

Simplifying

c = 1.166666667o-1s-2x-1 + -0.8333333333ino-1s-1

Reorder the terms:

c = -0.8333333333ino-1s-1 + 1.166666667o-1s-2x-1

Giải thích các bước giải:

Đáp án:

$\left\{ \begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi \\ x = \arcsin \dfrac{1}{3} + m2\pi \\ x = \pi - \arcsin \dfrac{1}{3} + m2\pi \end{array} \right.\left( {k,\,\,m \in Z} \right).$

Lời giải:

\(\begin{array}{l} 6{\cos ^2}x + 5\sin x - 7 = 0\\ \Leftrightarrow 6\left( {1 - {{\sin }^2}x} \right) + 5\sin x - 7 = 0\\ \Leftrightarrow 6 - 6{\sin ^2}x + 5\sin x - 7 = 0\\ \Leftrightarrow 6{\sin ^2}x - 5\sin x + 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = \dfrac{1}{2}\\ \sin x = \dfrac{1}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi \\ x = \arcsin \dfrac{1}{3} + m2\pi \\ x = \pi - \arcsin \dfrac{1}{3} + m2\pi \end{array} \right.\left( {k,\,\,m \in Z} \right). \end{array}\)

Vậy $\left\{ \begin{array}{l} x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi \\ x = \arcsin \dfrac{1}{3} + m2\pi \\ x = \pi - \arcsin \dfrac{1}{3} + m2\pi \end{array} \right.\left( {k,\,\,m \in Z} \right).$