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Form 4 mathematics trigonometry III questions and answers
Solve the equation 3cos x =#2sin^2x# where #0^0≤ x ≤ 360^0#
(6m 32s)
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1.
Solve the equation:Sin (#1/2x-30^0#) = cos x for 0 < x < #90^0#
2.
Solve for x: 4 sin #(x +20)^0# =3 for #0^0# = x = #360^0#
3.
Solve the equation;sin #5/2theta =1/2#for #0^0 = theta = 180^0#
4.
Solve for #theta# in the equation: #Sin (2theta – 10^0) #= -0.5 for #0^0 = theta = 360^0#
5.
Determine the amplitude and the period for the graph of y =6 sin #{ x/2 - 90}^0#
6.
Find all the values of #theta# between #0^0# and #360^0# satisfying the equation 5 sin #theta# = -4.
7.
Solve the equation;cos #(3theta+120^0)=frac{sqrt3}{2}# For # 0^0 = theta = 180^0#
8.
Solve the equation: #8s^2 + 2s -3 =0# Hence solve the equation: #8sin^2theta +2sintheta-3=0# For #0^0 = theta = 180^0#
9.
Solve the equation: #2sin^2#(x -#30^0#) = cos #60^0# For -#180^0 = x= 180^0#
10.
Given that sin (x +#30^0#) = cos 2x for #0^0#= x= #90^0# find the value of x. Hence find the value of #cos^2#3x.
11.
Solve the equation: #4Sin^2theta# + #4 Cos theta# = 5 for #0^0 = theta = 360^0#.Give the answer in degrees.
12.
Solve the equation: #3tan^2#x – 4 tan x – 4 =0. For #0^0 = x = 180^0#.
13.
Given that #x^0# is an angle in the first quadrant such that 8#sin^2x# +2cosx -5 =0 find; a) cos x b).tan x.
14.
Given that Cos 2x = 0.8070, find x when #0^0 = x = 180^0#
15.
Solve the equation 3cos x =#2sin^2x# where #0^0= x = 360^0#
16.
Solve the equation: 2#cos2theta# = 1 For #0^0= theta = 360^0#
17.
Solve the equation: Sin #(3x+30)^0# =#frac{sqrt3}{ 2}# ,for For #0^0 = x = 90^0#
18.
Solve:4 – 4 #cos^2alpha# = 4sin#alpha# -1 For #0^0 = alpha = 360^0#
19.
Solve the equation: Sin#(2t +10)^0# =0.5 for #0^0= t = 180^0#
20.
Given that sin#(x+20)^0# = -0.7660, find x to the nearest degree, for #0^0= x = 360^0#
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