Show that the equation 6cos2x=4−sinx6\cos^2 x = 4 - \sin x6cos2x=4−sinx can be written in the form 6sin2x−sinx−2=06\sin^2 x - \sin x - 2 = 06sin2x−sinx−2=0
Hence solve, for 0°⩽x<360°0° \leqslant x < 360°0°⩽x<360°, the equation 6cos2x=4−sinx6\cos^2 x = 4 - \sin x6cos2x=4−sinx, giving your answers to 1 decimal place where appropriate.
97 exam-style questions on WJEC A Level Maths 1.5 Trigonometry, covering 1.5.1 Trigonometry, 1.5.2 Trigonometry, 1.5.3 Trigonometry, 1.5.4 Trigonometry, and 1.5.5 Trigonometry. Each one has a worked solution and a mark scheme showing where the marks go.