18. Find the general solution of each of the following equations. \( \begin{array}{ll}\text { (a) } \sin 2 \theta=\cos \theta & \text { (b) } \sin 2 \theta=-\cos 2 \theta \\ \text { (c) } \sin 2 \theta=2 \cos \theta \\ \text { (d) } \sin 2 \theta=2 \cos 2 \theta & \text { (e) } \cos 2 \theta=4 \sin \theta+3\end{array} \) (f) \( \cos \left(4 x+10^{\circ}\right) \cos \left(20^{\circ}-3 x\right)-\sin \left(4 x+10^{\circ}\right) \sin \left(20^{\circ}-3 x\right)=\cos \left(2 x-60^{\circ}\right) \) (a) Prove that \( 4 \sin ^{2} \theta-11 \cos \theta=1 \) has no real solution for \( \theta \in\left[90^{\circ} ; 270^{\circ}\right] \)

To solve the equations and prove the statement, we will go through each part step by step. ### (a) Solve \( \sin 2\theta = \cos \theta \) Using the double angle identity, we have: \[ \sin 2\theta = 2 \sin \theta \cos \theta \] Thus, the equation becomes: \[ 2 \sin \theta \cos \theta = \cos \theta \] Assuming \( \cos \theta \neq 0 \), we can divide both sides by \( \cos \theta \): \[ 2 \sin \theta = 1 \implies \sin \theta = \frac{1}{2} \] The general solution for \( \sin \theta = \frac{1}{2} \) is: \[ \theta = 30^\circ + k \cdot 360^\circ \quad \text{or} \quad \theta = 150^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z} \] Now, consider the case when \( \cos \theta = 0 \): \[ \theta = 90^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z} \] Thus, the general solution for (a) is: \[ \theta = 30^\circ + k \cdot 360^\circ, \quad \theta = 150^\circ + k \cdot 360^\circ, \quad \theta = 90^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z} \] ### (b) Solve \( \sin 2\theta = -\cos 2\theta \) Using the double angle identity: \[ \sin 2\theta = -\cos 2\theta \implies \tan 2\theta = -1 \] The general solution for \( \tan 2\theta = -1 \) is: \[ 2\theta = 135^\circ + k \cdot 180^\circ \implies \theta = 67.5^\circ + k \cdot 90^\circ, \quad k \in \mathbb{Z} \] ### (c) Solve \( \sin 2\theta = 2 \cos \theta \) Using the double angle identity: \[ 2 \sin \theta \cos \theta = 2 \cos \theta \] Assuming \( \cos \theta \neq 0 \): \[ \sin \theta = 1 \implies \theta = 90^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z} \] If \( \cos \theta = 0 \): \[ \theta = 90^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z} \] Thus, the general solution for (c) is: \[ \theta = 90^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z} \] ### (d) Solve \( \sin 2\theta = 2 \cos 2\theta \) Using the double angle identity: \[ 2 \sin \theta \cos \theta = 2(1 - 2\sin^2 \theta) \] This simplifies to: \[ \sin \theta \cos \theta = 1 - 2\sin^2 \theta \] Rearranging gives: \[ 2\sin^2 \theta + \sin \theta \cos \theta - 1 = 0 \] This is a quadratic in \( \sin \theta \). Let \( x = \sin \theta \): \[ 2x^2 + x\sqrt{1-x^2} - 1 = 0 \] This requires solving for \( x \) and substituting back to find \( \theta \). ### (e) Solve \( \cos 2\theta = 4 \sin \theta + 3 \) Using the double angle identity: \[ 2\cos^2 \theta - 1 = 4 \sin \theta + 3 \] This simplifies to: \[ 2\cos^2 \theta - 4\sin \theta - 4 = 0 \] Substituting \( \cos^2 \theta = 1 - \sin^2 \theta \) leads to a quadratic in \( \sin \theta \). ### (f) Solve \( \cos(4x + 10^\circ) \cos(20^\circ - 3x) - \sin(4x + 10^\circ) \sin(20^\circ - 3x) = \cos(2x - 60^\circ) \) Using the cosine addition formula: \[ \cos(A + B) = \cos A \cos B - \sin A \sin B \] This gives: \[ \cos((4x + 10^\circ) + (20^\circ - 3x)) = \cos(2x - 60^\circ) \] Simplifying leads to: \[ \cos(x + 30^\circ) = \cos(2x - 60^\circ) \] This can be solved using the properties of cosine. ### Prove \( 4 \sin^2 \theta - 11 \cos \theta = 1 \) has no real solution for \( \theta \in [90^\circ, 270^\circ] \) Rearranging gives: \[ 4 \sin^2 \theta = 11 \cos \theta + 1 \] Using \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ 4(1 - \cos^2 \theta) = 11 \cos \theta + 1 \] This leads to: \[ 4 - 4\cos^2 \theta = 11 \cos \theta + 1 \] Rearranging gives: \[ 4\cos^2 \theta + 11 \cos \theta - 3 = 0 \] Using the discriminant: \[ D = 11^2 - 4 \cdot 4 \cdot (-3) = 121 + 48 = 169 \] The roots are: \[ \cos \theta = \frac{-11 \pm 13}{8} \implies \cos \theta = \frac{1}{4} \text{ or } \cos \theta = -3 \] Since \( \cos \theta = -3 \) is not possible, we only consider \( \cos \theta = \frac{1}{4} \). In the interval \( [90^\circ, 270^\circ] \), \( \cos \theta \) is negative, thus there are no real solutions in this interval. ### Summary of Solutions - (a)