Lanjutan 3 Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 16.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{75}^{\circ },{285}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{75}^{\circ },{343}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{75}^{\circ },{344}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{75}^{\circ },{345}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{75}^{\circ },{346}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}}\\ & \cos (x-30{)}^{\circ }=\cos {45}^{\circ }\\ & \iff x-{30}^{\circ }=\pm {45}^{\circ }+k{.360}^{\circ }\\ & \iff x={30}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={75}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{15}^{\circ }(tm)\\ & k=1\Rightarrow x_1={75}^{\circ }+{360}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{15}^{\circ }+{360}^{\circ }={345}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {75}^{\circ },{345}^{\circ }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }<x<{360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{100}^{\circ },{330}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{30}^{\circ },{330}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{120}^{\circ },{300}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{60}^{\circ },{120}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{50}^{\circ },{300}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos x^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ & \cos x^{\circ }=\cos {30}^{\circ }\\ & \iff x=\pm {30}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={30}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{30}^{\circ }(tm)\\ & k=1\Rightarrow x_1={30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{30}^{\circ }+{360}^{\circ }={330}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {30}^{\circ },{330}^{\circ }}\right\}\end{array}\end{array}$