6sin^2x+5cosx-2=0 sin2x=1-cos2x
To solve the equations you've provided, let's start with the first equation:
Equation 1: \( 6\sin^2 x + 5\cos x - 2 = 0 \)
Step 1: Use \( \sin^2 x = 1 - \cos^2 x \)
We can substitute \( \sin^2 x = 1 - \cos^2 x \) into the equation:
\[ 6(1 - \cos^2 x) + 5\cos x - 2 = 0 \]
Expanding the equation:
\[ 6 - 6\cos^2 x + 5\cos x - 2 = 0 \]
This simplifies to:
\[ -6\cos^2 x + 5\cos x + 4 = 0 \]
Step 2: Rearranging the equation
To make it a standard quadratic form, multiply through by -1:
\[ 6\cos^2 x - 5\cos x - 4 = 0 \]
Step 3: Using the quadratic formula
Let \( u = \cos x \). The equation is now \( 6u^2 - 5u - 4 = 0 \).
The quadratic formula is given by:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For \( a = 6, b = -5, c = -4 \):
\[ u = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 6 \cdot (-4)}}{2 \cdot 6} \]
Calculate the discriminant:
\[ 25 + 96 = 121 \]
Now substitute back into the equation:
\[ u = \frac{5 \pm \sqrt{121}}{12} \]
This gives us:
\[ u = \frac{5 \pm 11}{12} \]
Calculating the two possible values for \( u \):
\[ u_1 = \frac{16}{12} = \frac{4}{3} \quad (\text{not valid since } -1 \leq \cos x \leq 1) \] \[ u_2 = \frac{-6}{12} = -\frac{1}{2} \]
Step 4: Solve \( \cos x = -\frac{1}{2} \)
The angle(s) which satisfy \( \cos x = -\frac{1}{2} \) are:
\[ x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{4\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
Equation 2: \( \sin 2x = 1 - \cos 2x \)
Using the double angle identities, call \( y = 2x \). Then we can rewrite the equation:
\[ \sin y = 1 - \cos y \]
Also using the identity \( \cos y = \sqrt{1 - \sin^2 y} \), we can manipulate as follows:
\[ \sin y + \cos y = 1 \]
Squaring both sides:
\[ \sin^2 y + 2\sin y \cos y + \cos^2 y = 1 \]
Using \( \sin^2 y + \cos^2 y = 1 \):
\[ 1 + 2\sin y \cos y = 1 \]
This simplifies to:
\[ 2\sin y \cos y = 0 \]
This means either \( \sin y = 0 \) or \( \cos y = 0 \).
Step 5: Solve \( \sin y = 0 \)
\[ y = n\pi \quad (n \in \mathbb{Z}) \]
Step 6: Solve \( \cos y = 0 \)
\[ y = \frac{\pi}{2} + m\pi \quad (m \in \mathbb{Z}) \]
Substituting back \( y = 2x \):
- For \( sin(y) = 0 \):
\[ 2x = n\pi \implies x = \frac{n\pi}{2} \]
- For \( cos(y) = 0 \):
\[ 2x = \frac{\pi}{2} + m\pi \implies x = \frac{\pi}{4} + \frac{m\pi}{2} \]
Summary:
The solutions are:
From \( 6\sin^2 x + 5\cos x - 2 = 0 \):
- \( x = \frac{2\pi}{3} + 2k\pi \) or \( x = \frac{4\pi}{3} + 2k\pi \)
From \( \sin 2x = 1 - \cos 2x \):
- \( x = \frac{n\pi}{2} \)
- \( x = \frac{\pi}{4} + \frac{m\pi}{2} \)
Combine both sets of solutions from the respective equations.
