User interface language: English | Español

Date May 2021 Marks available 1 Reference code 21M.1.SL.TZ2.3
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Show that Question number 3 Adapted from N/A

Question

Show that the equation 2cos2x+5sinx=4 may be written in the form 2sin2x-5sinx+2=0.

[1]
a.

Hence, solve the equation 2cos2x+5sinx=4, 0x2π.

[5]
b.

Markscheme

METHOD 1

correct substitution of cos2x=1-sin2x            A1

21-sin2x+5sinx=4

2sin2x-5sinx+2=0          AG

 

METHOD 2

correct substitution using double-angle identities             A1

2cos2x-1+5sinx=3

1-2sin2x-5sinx=3

2sin2x-5sinx+2=0           AG

 

[1 mark]

a.

EITHER

attempting to factorise              M1

(2sinx1)(sinx2)                   A1

 

OR

attempting to use the quadratic formula            M1

sinx=5±52-4×2×24=5±34         A1

 

THEN

sinx=12           (A1)

x=π6,5π6                  A1A1

 

[5 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.6—Pythagorean identity, double angles
Show 23 related questions
Topic 3— Geometry and trigonometry » SL 3.8—Solving trig equations
Topic 3— Geometry and trigonometry

View options