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 \( y = 2x \): 1. For \( sin(y) = 0 \): \[ 2x = n\pi \implies x = \frac{n\pi}{2} \] 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.